Here's what I perceive to be a mathematically and logically precise presentation of the theorem, let me know if this helps.
Mathematical Preliminaries
First let me introduce some precise notation so that we don't encounter any issues with "infinitesimals" etc. Given a field $\phi$, let $\hat\phi(\alpha, x)$ denote a smooth one-parameter family of fields for which $\hat \phi(0, x) = \phi(x)$. We call this family a deformation of $\phi$ (in a previous version I called this a "flow"). Then we can define the variation of $\phi$ under this deformation as the first order approximation to the change in $\phi$ as follows:
Definition 1. (Variation of field)
$$
\delta\phi(x) = \frac{\partial\hat\phi}{\partial\alpha}(0,x)
$$
This definition then implies the following expansion
$$
\hat\phi(\alpha, x) = \phi(x) + \alpha\delta\phi(x) + \mathcal O(\alpha^2)
$$
which makes contact with the notation in many physics books like Peskin and Schroeder.
Note: In my notation, $\delta\phi$ is NOT an "infinitesimal", it's the coefficient of the parameter $\alpha$ in the first order change in the field under the deformation. I prefer to write things this way because I find that it leads to a lot less confusion.
Next, we define the variation of the Lagrangian under the deformation as the coefficient of the change in $\mathcal L$ to first order in $\alpha$;
Definition 2. (Variation of Lagrangian density)
$$
\delta\mathcal L(\phi(x), \partial_\mu\phi(x)) = \frac{\partial}{\partial\alpha}\mathcal L(\hat\phi(\alpha, x), \partial_\mu\hat\phi(\alpha, x))\Big|_{\alpha=0}
$$
Given these definitions, I'll leave it to you to show
Lemma 1.
For any variation of the fields $\phi$, the variation of the Lagrangian density satisfies
\begin{align}
\delta\mathcal L
&= \left(\frac{\partial \mathcal L}{\partial\phi} - \partial_\mu\frac{\partial\mathcal L}{\partial(\partial_\mu\phi)}\right)\delta\phi + \partial_\mu K^\mu,\qquad K^\mu = \frac{\partial\mathcal L}{\partial(\partial_\mu\phi)}\delta\phi
\end{align}
You'll need to use (1) The chain rule for partial differentiation, (2) the fact $\delta(\partial_\mu\phi) = \partial_\mu\delta\phi$ which can be proven from the above definition of $\delta\phi$ and (3) the product rule for partial differentiation.
Noether's theorem in steps
Let a particular flow $\hat\phi(\alpha, x)$ be given.
Assume that for this particular deformation, there exists some vector field $J^\mu\neq K^\mu$ such that
$$
\delta\mathcal L = \partial_\mu J^\mu
$$
Notice, that for any field $\phi$ that satisfies the equation of motion, Lemma 1 tells us that
$$
\delta \mathcal L = \partial_\mu K^\mu
$$
Define a vector field $j^\mu$ by
$$
j^\mu = K^\mu - J^\mu
$$
Notice that for any field $\phi$ satisfying the equations of motion steps 2+ 3 + 4 imply
$$
\partial_\mu j^\mu = 0
$$
Q.E.D.
Important Notes!!! If you follow the logic carefully, you'll see that $\delta \mathcal L = \partial_\mu K^\mu$ only along the equations of motion. Also, part of the hypothesis of the theorem was that we found a $J^\mu$ that is not equal to $K^\mu$ for which $\delta\mathcal L = \partial_\mu J^\mu$. This ensures that $j^\mu$ defined in the end is not identically zero! In order to find such a $J^\mu$, you should not be using the equations of motion. You should be applying the given deformation to the field and seeing what happens to it to first order in the "deformation parameter" $\alpha$.
Addendum. 2020-07-02 (Free scalar field example.)
A concrete example helps clarify the theorem and the remarks made afterward. Consider a single real scalar field $\phi:\mathbb R^{1,3}\to\mathbb R$. Let $m\in\mathbb R$ and $\xi\in\mathbb R^{1,3}$, and consider the following Lagrangian density and deformation (often called spacetime translation):
$$
\mathcal L(\phi, \partial_\mu\phi) = \frac{1}{2}\partial_\mu\phi\partial^\mu\phi - \frac{1}{2}m^2\phi, \qquad \hat\phi(\alpha, x) = \phi(x + \alpha\xi)
$$
Computation using the definition of $\delta\mathcal L$ (plug the deformed field into $\mathcal L$, take the derivative with respect to $\alpha$, and set $\alpha = 0$ at the end) but without ever invoking the equation of motion (Klein-Gordon equation) for the field gives
$$
\delta \mathcal L = \partial_\mu(\xi^\nu\delta^\mu_\nu \mathcal L), \qquad \frac{\partial\mathcal L}{\partial(\partial_\mu\phi)}\delta\phi = \xi^\nu\partial_\nu\phi\partial^\mu\phi
$$
It follows that
$$
J^\mu = \xi^\nu\delta^\mu_\nu \mathcal L, \qquad K^\mu = \xi^\nu\partial_\nu\phi\partial^\mu\phi
$$
and therefore
$$
j^\mu = \xi^\nu(\partial_\nu\phi\partial^\mu\phi -\delta^\mu_\nu\mathcal L)
$$
If e.g. one chooses $\tau > 0$ and sets $\xi = (\tau, 0, 0, 0)$, then the deformation is time translation, and conservation of $j^\mu$ yields conservation of the Hamiltonian density associated with $\mathcal L$ as the reader can check.
Suppose, instead, that in the process of computing $\delta \mathcal L$, one were to further invoke the following equation of motion which is simply the Euler-Lagrange equation for the Lagrangian density $\mathcal L$:
$$
\partial^\mu\partial_\mu\phi = -m^2\phi,
$$
Then one finds that
$$
\delta\mathcal L = \partial_\mu(\xi^\nu\partial_\nu\phi\partial_\mu\phi)
$$
so $J^\mu = K^\mu$ and therefore $j^\mu = 0$, which is uninformative.
A translation by $x^\nu \to x^\nu - \epsilon^\nu$ corresponds to an infinitesimal transformation of the fields, by
$$\phi \to \phi + \epsilon^\nu \partial_\nu \phi$$
as we are performing an active rather than passive transformation. The Lagrangian transforms as,
$$\mathcal{L}\to \mathcal{L}+\epsilon^\nu \partial_\nu \mathcal{L}$$
by substituting $\phi$ into the Lagrangian. Notice the change is up to a total derivative, and hence Noether's theorem is applicable to the symmetry. The conserved current density is given by,
$$j^\mu = \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}X(\phi)-F^\mu(\phi)$$
where $X=\delta\phi$ and $F^\mu$ is such that $\partial_\mu F^\mu=\delta \mathcal{L}$ infinitesimally. For our case, we obtain the symmetric stress-energy tensor (analogous to that of general relativity),
$$T^\mu_\nu=\frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)} \partial_\nu \phi - \delta^\mu_\nu \mathcal{L}$$
where the Kronecker delta is raised with the Minkowski metric. The current satisfies, $\partial_\mu T^{\mu}_\nu = 0$, and the corresponding Noether charge,
$$E=\int \mathrm{d}^3 x \, T^{00}$$
is the total energy of the system, whereas,
$$P^i = \int \mathrm{d}^3 x \, T^{0i}$$
is the $i$th component of the total momentum of the field, where $i=(x,y,z)$ only. A caveat: the stress-energy tensor derived by Noether's theorem is not always symmetric, and may require the addition of a term which satisfies the continuity equation, and ensures symmetry in the indices.
Alternate Method
Recall to obtain the Einstein field equations in general relativity, we may vary the Einstein-Hilbert action,
$$S\sim \int \mathrm{d}^4 x \, \sqrt{-g} \, \left( R + \mathcal{L}\right)$$
Similarly, in quantum field theory, we may promote our Minkowski metric to a generic metric tensor, thereby replacing the kinetic term of the Lagrangian with covariant derivatives. Up to some constants, the stress-energy tensor is given by
$$T^{\mu\nu} \sim \frac{1}{\sqrt{-g}} \frac{\partial (\sqrt{-g}\mathcal{L})}{\partial g^{\mu\nu}}$$
evaluated at $g_{\mu\nu}=\eta_{\mu\nu}$, which is precisely the definition we implement when obtaining the Einstein field equations for general relativity.
Best Answer
First of all, let's see what Noether's Theorem says about your specific case (Klein-Gordon under global rescaling of the fields). Noether's theorem states that
The current in object is given by
$$ J^{\mu}=-T_{\nu}^{\mu}\ \delta x^{\nu}+\frac{\partial \mathcal{L}}{\partial \phi^{a}_{,\mu}}\ \delta \phi^{a} $$ where the $\phi^{a}$ are the fields whose dynamics is described by the action, $T^{\mu}_{\nu}$ is the canonical energy-momentum tensor of the theory and $\delta x^{\nu}$ and $\delta \phi^{a}$ are the infinitesimal generators of the symmetry. In your case,
$$ \phi\to e^{\epsilon}\phi\approx (1+\epsilon)\phi $$ so that $\delta \phi=\epsilon \phi$. (If your $\alpha$ is negative, then the symmetry is not differentiable in the Noetherian sense, as there is no infinitesimal generator. In fact, rescaling by a factor of $-1$ is part of the discrete, non differentiable, multiplicative group $\{+1,-1\}$). Then $$ J^{\mu}=\epsilon\ \phi\partial^{\mu}\phi $$ But there is no reason why this Noetherian current should be conserved. In fact, removing the $\epsilon$ from the above expression, we see that the divergence is proportional to the Lagrangian, $$ \partial_{\mu}(J^{\mu}/\epsilon)=\partial_{\mu}(\phi\partial^{\mu}\phi)=\phi\ \partial_{\mu}\partial^{\mu}\phi+\partial_{\mu}\phi\partial^{\mu}\phi=\partial_{\mu}\phi\partial^{\mu}\phi-m^{2}\phi^{2}=2\mathcal{L} $$ where in the third identity I used the equations of motion $\partial^{2}\phi=-m^{2}\phi$. The most general solution to Klein-Gordon's equation is given by $$ \phi(x)=\int\frac{d^{3}k}{(2\pi)^{3}}\ \Big\{e^{-ik_{\mu}x^{\mu}}\ A(\vec{k})+e^{+ik_{\mu}x^{\mu}}\ B(\vec{k})\Big\} $$ with $k_{\mu}k^{\mu}=m^{2}$, but not even in the plane-wave case (say, $\phi(x)=e^{-ik_{\mu}x^{\mu}}$) the Lagrangian is zero: $$ 2\mathcal{L}[e^{-ik_{\mu}x^{\mu}}]=-k_{\mu}k^{\mu}\ e^{-2ik_{\mu}x^{\mu}}-m^{2}\ e^{-2ik_{\mu}x^{\mu}}=-2m^{2}\ e^{-2ik_{\mu}x^{\mu}}\neq0 $$ for $m\neq 0$. The fact that it is the mass that determines whether $J^{\mu}$ is conserved or not is not accidental, as we will see later on.
$$ $$
Now, you ask whether there is any theorem, analogous to Noether's theorem, that allows you to derive conserved currents from a generalized concept of "symmetry under some transformation". Specifically, you ask for a theorem that does so with symmetries of the Euler-Lagrange equations, rather than of the action (as a matter of fact, your transformation doesn't leave the action invariant, it multiplies it by a factor of $\alpha$). I can't really say that such a theorem does not exist, but I can safely say that I don't know of any, and that I doubt that such a theorem can, in general, be true. Here is why. The intuition behind Noether's theorem is that the values of the fields that solve the minimization problem can, by definition, be shifted by an infinitesimal amount without changing the value of action (meaning that the functional derivative of the action with respect to the "variation field" is zero). Then you can ask what happens to the action if such a transformation is made on extremal fields, and you find that to the shift of the fields there corresponds a shift of the action given by
$$ \delta S=\int_{\Omega} d^{d}x\ \partial_{\mu}J^{\mu}\qquad\qquad(\star) $$ (the equations of motion being satisfied by hypothesis) where $J^{\mu}$ is again Noether's current and $\Omega$ is an arbitrary domain of integration. Then you conclude that if the action is invariant with respect to the transformation, you should have $\delta S=0$, so that $\partial_{\mu}J^{\mu}=0$. Note that the hypothesis of the invariance of the action is brought up only at the end: it is an additional hypothesis, independent of the result $(\star)$. The latter is an identity that comes about under the only hypothesis that the fields (1) solve Euler-Lagrange's equations (2) are shifted by an infinitesimal amount. No assumption about the nature of the transformations or their being part of a group that leaves invariant the action is made. So $(\star)$ holds even when the transformation multiplies the action by a factor of $\alpha>0$, and we can use it to derive the following result. As $$ S\to \alpha S= e^{\epsilon} S\approx(1+\epsilon) S\qquad\Longrightarrow\qquad \delta S=\epsilon S $$ we have that, under such a transformation $$ \int_{\Omega} d^{d}x\ \partial_{\mu}J^{\mu}=\epsilon S $$ so we can conclude that, in general, Noether's current is not divergence-free under such an action-rescaling transformation (as we have seen through the example of the Klein-Gordon action). I emphasized "Noether's" because of course there may be some other current that is divergence-free instead of the standard Noether's one. But if one removes the hypothesis of the invariance of the action under some symmetry, there is little left to say about the symmetries of the solutions of Euler-Lagrange's equations: the connection between symmetries and conservation lies in the very fact that the solutions themselves (before even thinking about symmetries) are such that an infinitesimal variation on the solving values leaves the action invariant . Then the hypothesis that under the given transformation the action gets rescaled is somewhat incompatible with the essential property of the solutions, i.e. with the Euler-Lagrange's equations themselves. This is why I find it difficult to believe that such a theorem would, in general, be true.
To end this answer, I want to mention two more things. First of all, the fact that in Klein-Gordon's case the Euler-Lagrange's equation are invariant under a constant rescaling of the fields follows from the fact that the Lagrangian is quadratic in the fields. Any such Lagrangian always gives rise to linear Euler-Lagrange's equation, which in turn are always symmetric under constant rescalings. The same holds for Dirac's Lagrangian and for Yang-Mills's Lagrangian (for free gauge bosons). Second of all, there is indeed a scaling transformation that leaves the Action invariant in the sense of Noether's. Consider making the transformation $$ x^{\mu}\to e^{\epsilon}\ x^{\mu}\qquad\qquad \phi\to e^{\epsilon}\ \phi $$ then $$ \partial_{\mu}\to e^{-\epsilon}\ \partial_{\mu} $$ and we see that, given $m=0$, Klein-Gordon's Action is invariant under such a transformation. The latter is called a "(constant) conformal transformation", and the corresponding Noether's current $$ j^{\mu}=J^{\mu}/\epsilon=-T^{\mu}_{\nu}\ x^{\nu}+\phi\ \partial^{\mu}\phi $$ is, as the theorem proves, divergence-free. An analogous statement can be made for the Dirac and Yang-Mills massless Lagrangians. Now, we have
$$ \partial_{\mu}j^{\mu}=-\partial_{\mu}(T^{\mu}_{\nu}\ x^{\nu})+\partial_{\mu}(\phi\ \partial^{\mu}\phi) $$ As we know from translational invariance that $\partial_{\mu} T^{\mu}_{\nu}=0$, and given the calculation we made before, $$ \partial_{\mu}j^{\mu}=-T^{\mu}_{\nu}\ \delta^{\nu}_{\mu}+2\mathcal{L} $$ Let's calculate $T^{\mu}_{\nu}\ \delta^{\nu}_{\mu}=T^{\mu}_{\mu}$ for the massless Klein-Gordon Lagrangian. We have $$ T^{\mu}_{\mu}=\frac{\partial\mathcal{L}}{\partial \phi_{,\mu}}\ \partial_{\mu}\phi-\mathcal{L}\ \delta^{\mu}_{\mu}=\partial^{\mu}\phi\partial_{\mu}\phi-d\mathcal{L}=(2-d)\ \mathcal{L} $$ where $d$ is the dimension of spacetime ($d=4$ in the standard theory). Then $$ \partial_{\mu}j^{\mu}=d\ \mathcal{L} $$ so this divergence ($\partial_{\mu}j^{\mu}$) is $d/2$ times the divergence (let's call it $\partial_{\mu}j'^{\,\mu}$) you get from the transformation that you proposed in your question. This explains why for the latter we found $$ \partial_{\mu}j'^{\,\mu}\propto m^{2} $$ If $m=0$, then your transformation can be extended to a conformal transformation which is a true symmetry of Klein-Gordon's action, such that the Noether's current associated to it is conserved.