Dear Jenkins, the theories you want to construct are "noncritical string theories" and they're less interesting and less consistent than the "critical string theories".
First, the Nambu-Goto action - the proper area of the world sheet - is nonlinear. It includes square roots etc. It's much better to introduce an auxiliary metric tensor on the world sheet and the action for the coordinates $X$ becomes nice and bilinear - a free theory.
However, we don't want new degrees of freedom to be added. The 2D metric tensor has three independent components. Two of them may be set to a standard form by the 2 degrees of freedom in the 2D coordinate reparameterization symmetry; and the third by the Weyl symmetry if it exists.
If it doesn't exist, it's too bad. The auxiliary world sheet metric may only be brought to the form of $e^\phi \eta_{ab}$. That means that $\phi$, determining the overall scaling, becomes another function of the world sheet coordinates $(\sigma,\tau)$, very analogously to the spacetime coordinates $X(\sigma,\tau)$. In fact, it is really valid to say that the parameter determining the overall scaling of the metric is another spacetime coordinate.
If this coordinate were totally identical to the other coordinates, then there would also be a translation symmetry in the $\phi$ direction - but that's equivalent to the Weyl symmetry (multiplicative scaling of $e^\phi$ is the same thing as additive shifts to $\phi$). Because by assumption, the Weyl symmetry doesn't hold in your theory, the new spacetime coordinate $\phi$ can't have quite the same properties as the other spacetime coordinates.
However, in normal circumstances, you obtain the violations of the Weyl invariance as a disease. In particular, if you try to study string theory in a non-critical dimension, i.e. $D\neq 26$ or $D\neq 10$, you will find out that the field $\phi$ doesn't decouple and the path integral, when calculated including the one-loop accuracy, still depends on $\phi$. So the Weyl symmetry, equivalent to an additive shift of $\phi$ by a function of the world sheet, is not a symmetry.
As I said, this can be interpreted as $\phi$'s becoming a new spacetime coordinate. But if you try to calculate the effective action in the new spacetime that has an additional dimension $\phi$, you will find out that the laws of physics are not invariant under translations in $\phi$ - that's nothing else than the failure of the theory to be Weyl-invariant.
In particular, you will find out that the dilaton linearly depends on $\phi$: search for papers about "linear dilaton". The squared gradient of the dilaton is related to the surplus or excess (if it is time-like or space-like) of the spacetime coordinates, relatively to the critical dimension.
If the spacetime has two dimensions, one may choose the dilaton to depend on the (only) spacelike coordinate $\phi=X^1$ in such a way that the theory including $\phi$ is Weyl-invariant again. In this case, it's useful to consider not only the right linear dilaton - solving the equations of motion - but also a non-trivial background for the tachyon. One ends up with the so-called "Liouville theory" - a "linear dilaton" theory with some extra tachyonic profile in a non-critical stringy $D=2$ spacetime - which is slightly more consistent than other noncritical string theories. The Liouville theory may also be described by a quantum mechanical model with a large matrix - the old matrix theory.
I think the original source of this claim is the famous unpublished paper of Luescher and Mack. Everyone's citing it. It is more rigorous mathematically and general (they don't assume parity) than Di Francesco. It starts on pages 1-2 of the manuscript. The proof below is basically the same proof, just with added details and a little bit different notation.
Assumptions: $T_{\mu\nu}$ are local covariant fields of dimension 2 and $$T^{\dagger}_{\mu\nu} = T_{\mu\nu}, \quad T_{\mu\nu} = T_{\nu\mu},\quad \partial^{\mu}T_{\mu\nu} = 0.$$ Moreover, one assumes that the correlation functions are well-defined and behaved, and satisfy usual axioms of CFT as stated e.g. in this page of nLab.
Proof:
First we define the 2-point functions
$$
S_{\mu\nu\rho\sigma}(\vec{x}, \vec{y}) := \text{analytic continuation of } \langle \Omega \mid T_{\mu\nu}(\vec{x}) T_{\rho\sigma}(\vec{y}) \mid \Omega \rangle.
$$
All $S$'s are translationally invariant, so we may as well set $\vec{y}=0$
so that we are left with
$$
S_{\mu\nu\rho\sigma}(\vec{x}) := \langle \Omega \mid T_{\mu\nu}(\vec{x}) T_{\rho\sigma}(0) \mid \Omega \rangle.
$$
This function is real analytic for $\vec{x}\in\mathbb{R}^2,\, \vec{x}\neq 0$, and because of invariance under dilations, and the assumption that $T_{\mu\nu}$'s are of conformal weight 2, we have
\begin{equation}
S_{\mu\nu\rho\sigma}(\vec{x})=\lambda^4 S_{\mu\nu\rho\sigma}(\lambda \vec{x})\quad\forall\lambda\in\mathbb{R}.\quad\quad (1)
\end{equation}
Moreover, from $T_{\mu\nu}=T_{\nu\mu}$ we get
$$
S_{\mu\nu\rho\sigma}(\vec{x})=S_{\nu\mu\rho\sigma}(\vec{x})=S_{\mu\nu\sigma\rho}(\vec{x}).\quad\quad (2)
$$
Furthermore, by locality, invariance under translations and Equation 1 with $\lambda=-1$ we get
$$
S_{\mu\nu\rho\sigma}(\vec{x},0)=S_{\rho\sigma\mu\nu}(0,\vec{x})=S_{\rho\sigma\mu\nu}(-\vec{x},0)=S_{\rho\sigma\mu\nu}(\vec{x},0),
$$
or in shorthand notation
$$
S_{\mu\nu\rho\sigma}(\vec{x}) = S_{\rho\sigma\mu\nu}(\vec{x}) .\quad\quad (3)
$$
From Equations 2 and 3 we see that $S_{\mu\nu\rho\sigma}(\vec{x})$ has 6 independent components and also taking into account Equation 1 we can write
$$
S_{\mu\nu\rho\sigma}(\vec{x}) = (\vec{x}^{\,2})^{-4}\sum_{i=1}^6 \alpha_i f^{\,i} _{\mu\nu\sigma\rho}(\vec{x}),\quad \alpha_i\in\mathbb{C},\quad\quad (4)
$$
where
\begin{align}
&f^{1}_{\mu\nu\rho\sigma}(\vec{x})=(\vec{x}^{\,2})^2 g_{\mu\nu} g_{\rho\sigma},\\\\ &f^{2}_{\mu\nu\rho\sigma}(\vec{x})=(\vec{x}^{\,2})^2 (g_{\mu\rho} g_{\nu\sigma}+g_{\mu\sigma}g_{\nu\rho}),\\\\ &f^{3}_{\mu\nu\rho\sigma}(\vec{x})=\vec{x}^{\,2} (g_{\mu\nu} x_\rho x_\sigma + g_{\rho\sigma} x_\mu x_\nu ),\\\\ &f^{4}_{\mu\nu\rho\sigma}(\vec{x})=x_\mu x_\nu x_\rho x_\sigma,\\\\ &f^{5}_{\mu\nu\rho\sigma}(\vec{x})= \vec{x}^{\,2} \left (g_{\mu\nu} (x_\rho \varepsilon_{\sigma\delta}x^\delta+x_\sigma\varepsilon_{\rho\delta}x^\delta )+ g_{\rho\sigma} (x_\mu\varepsilon_{\nu\delta} x^\delta + x_\nu \varepsilon_{\mu\delta}x^\delta) \right),\\\\ &f^{6}_{\mu\nu\rho\sigma}(\vec{x})= \left( x_\mu \varepsilon_{\nu\delta}x^\delta +x_\nu \varepsilon_{\mu\delta}x^\delta \right) x_\rho x_\sigma + x_\mu x_\nu \left( x_\rho \varepsilon_{\sigma\delta}x^\delta + x_\sigma \varepsilon_{\rho\delta}x^\delta \right),
\end{align}
and
$$
g_{\mu\nu}=\delta_{\mu\nu},\quad \varepsilon_{\mu\nu}=-\varepsilon_{\nu\mu},\quad \varepsilon_{01}=+1,\quad \vec{x}^{\,2}=(x^0)^2+(x^1)^2.
$$
The continuity equation $\partial^\mu T_{\mu\nu}=0$ implies that $\partial^\mu S_{\mu\nu\rho\sigma}(\vec{x}) = 0$ and thus reduces the number of independent $\alpha_i$'s to 2 arbitrary constants $\alpha_+$ and $\alpha_-$ (checked myself using Mathematica):
\begin{alignat*}{6}
&\alpha_1 = 3 \alpha_+,\quad &&\alpha_2 = -\alpha_+,\quad \alpha_3 = -4\alpha_+,\quad \alpha_4 = 8\alpha_+,&\\
&\alpha_5 = \alpha_-,\quad &&\alpha_6 = -2\alpha_-. &
\end{alignat*}
Inserting these values into Equation 4 we obtain:
$$
S^{\mu}_{\;\mu\rho\sigma} (\vec{x}) = S^{\mu\;\rho}_{\;\mu\;\rho} (\vec{x}) = 0,
$$
i.e. $\langle\Omega \mid T^\mu_{\;\mu}(\vec{x})\, T^\rho_{\;\rho} (0)\mid\Omega\rangle = 0$, which by the Reeh-Schlieder Theorem implies that $T^\mu_{\;\mu}(\vec{x})=0$.
Now to be able to apply Reeh-Schlieder, we actually have to go to $\vec{x}=0$ as you have correctly pointed out. And the reason why we can do this is analytic continuation, citing p 110 of R. JOST: The General Theory of Quantized Fields, 1965 (the same step in a related lemma):
"Such a Wigtman distribution thus vanishes in the real regularity points and thus, by analytic continuation identically".
tl;dr analytic continuation
Best Answer
The (Belinfante-Rosenfeld) stress energy momentum tensor is defined as
$$T^{\mu\nu}\propto \frac{1}{\sqrt{-g}} \frac{\delta S}{\delta g_{\mu\nu}}$$
where the worldsheet metric is $g_{\mu\nu}$. By definition of the functional derivative, for any variation $\delta g_{\mu\nu}$, we have
$$\delta S = \int \frac{\delta S}{\delta g_{\mu\nu}} \delta g_{\mu\nu}$$
Consider now the case where $\delta g_{\mu\nu}$ is an infinitesimal Weyl invariance, or
$\delta g _{\mu\nu} = \Omega^2 g_{\mu\nu}$, where $\Omega$ is any function.
Weyl invariance of $S$ means that $\delta S$ has to vanish for all $\delta g_{\mu\nu}$ of this form, or
$$0=\int \frac{\delta S}{\delta g_{\mu\nu}}\Omega^2 g_{\mu\nu}$$
Then, the fundamental lemma of the calculus of variations implies tracelessness of the functional derivative $\frac{\delta S}{\delta g_{\mu\nu}}$, which is equal to the stress energy tensor up to various proportionality factors.
Incidentally, this line of reasoning also gives you things like Bianchi identities (try this for the Einstein action).