You probably made a mistake when you computed the curvature of the scaled manifold. In local coordinates $\mathrm{Riem} \approx \partial \Gamma + \Gamma^2$ and $\Gamma \approx g^{-1} \partial g$. So the Riemann curvature as a $(1,3)$ tensor is scale invariant. This means that the sectional curvature
$$ K(u,v) = \frac{\langle \mathrm{Riem}(u,v)v,u\rangle}{\langle u,u\rangle\langle v,v\rangle - \langle u,v\rangle^2} $$
scales like $g * g^{-2}$ or that $\tilde{K} = \lambda^{-1} K$ if $\tilde{g} = \lambda g$.
Your equation is incorrect, the correct condition is
$$\frac{1}{\kappa^2} + \left(\frac{\dot{\kappa}}{\tau\kappa^2}\right)^2 = \text{constant}$$
I will only show the $\Leftarrow$ part here.
Let $s$ be the arc length parametrization and $\vec{t}(s), \vec{n}(s), \vec{b}(s)$ be the
vectors appear in Frenet Serret equations. Define
$$\vec{\beta}(s) = \vec{\alpha}(s) + \frac{1}{\kappa(s)}\vec{n}(s) - \frac{\dot{\kappa}(s)}{\tau(s)\kappa(s)^2} \vec{b}(s)\tag{*1}$$
Differentiate it with respect to $s$, we get:
$$\begin{align}
\frac{d}{ds}\vec{\beta}(s)
= & \vec{t} - \frac{\dot{\kappa}}{\kappa^2}\vec{n} + \frac{1}{\kappa}(-\kappa \vec{t} + \tau \vec{b} )
- \frac{d}{ds}\left(\frac{\dot{\kappa}}{\tau\kappa^2}\right)\vec{b} -\frac{\dot{\kappa}}{\tau\kappa^2}(-\tau\vec{n})\\
= & \left(\frac{\tau}{\kappa} -\frac{d}{ds}\left(\frac{\dot{\kappa}}{\tau\kappa^2}\right)\right) \vec{b}\\
= & \frac{\tau\kappa^2}{\dot{\kappa}}\left(\frac{\dot{\kappa}}{\kappa^3} - \frac{\dot{\kappa}}{\tau\kappa^2}\frac{d}{ds}\left(\frac{\dot{\kappa}}{\tau\kappa^2}\right) \right) \vec{b}\\
= & -\frac{\tau\kappa^2}{2\dot{\kappa}}\frac{d}{ds}\left(\frac{1}{\kappa^2} + \left(\frac{\dot{\kappa}}{\tau\kappa^2}\right)^2\right) \vec{b}\\
= & \vec{0}
\end{align}$$
This implies $\vec{\beta}(s) = \vec{\beta}(0)$ is a constant. From this, we get
$$
\vec{\alpha} - \vec{\beta}(0)
= -\frac{1}{\kappa}\vec{n} + \frac{\dot{\kappa}}{\tau\kappa^2} \vec{b}
\quad\implies\quad \left|\vec{\alpha} - \vec{\beta}(0)\right|^2
= \frac{1}{\kappa^2} + \left(\frac{\dot{\kappa}}{\tau\kappa^2}\right)^2
= \text{constant}.
$$
i.e $\vec{\alpha}(s)$ lies on a sphere with $\beta(0)$ as center.
Motivation of above proof
You may wonder how can anyone figure out the magic formula in $(*1)$. If you work out the
$\Rightarrow$ part of the proof where $\alpha(s)$ lies on a sphere centered at $\vec{c}$,
you should obtain a bunch of dot products between $\vec{\alpha}(s) - \vec{c}$ and $\vec{t}(s)$, $\vec{n}(s)$ and $\vec{b}(s)$. In particular, you should get:
$$\begin{cases}
\vec{t} \cdot (\vec{\alpha} - \vec{c}) & = 0\\
\vec{n} \cdot (\vec{\alpha} - \vec{c}) & = -\frac{1}{\kappa}\\
\vec{b} \cdot (\vec{\alpha} - \vec{c}) &= \frac{\dot{\kappa}}{\tau\kappa^2}
\end{cases}$$
Using these, you can express the center $\vec{c}$ in terms of $\kappa, \tau$ like what we have in $(*1)$. If the curve does lie on a sphere, then the "center" should not move as $s$ changes. The proof of the $\Leftarrow$ part above is really using the given condition to verify the "center" so defined doesn't move.
Best Answer
Suppose $\alpha(s)$ is a unit speed curve lying in the sphere of radius $R$ centered at the origin. Then
$\alpha(s) \cdot \alpha (s) = R^2, \tag 1$
whence
$\dot \alpha(s) \cdot \alpha(s) = 0; \tag 2$
since
$\dot \alpha(s) = T(s), \tag 3$
the unit tangent vector to $\alpha(s)$, (2) becomes
$T(s) \cdot \alpha (s) = 0; \tag 4$
differentiating this equation yields
$\dot T(s) \cdot \alpha(s) + T(s) \cdot \dot \alpha(s) = 0; \tag 5$
we now recall (3), viz.
$\dot \alpha(s) = T(s) \tag 6$
and the Frenet-Serret equation
$\dot T(s) = \kappa(s) N(s); \tag 7$
then (5) yields
$\kappa(s) N(s) \cdot \alpha(s) + T(s) \cdot T(s) = 0; \tag 8$
also,
$T(s) \cdot T(s) = 1, \tag 9$
$T(s)$ being a unit vector. (8) may now be written
$\kappa(s) N(s) \cdot \alpha(s) = -1; \tag{10}$
note this forces
$\kappa(s) \ne 0; \tag{11}$
taking absolute values in (10) we find
$\kappa(s) \vert N(s) \cdot \alpha(s) \vert = 1; \tag{12}$
by Cauchy-Schwarz,
$ \vert N(s) \cdot \alpha(s) \vert \le \vert \alpha(s) \vert \vert N(s) \vert = R, \tag{13}$
since
$\vert \alpha(s) \vert = R \tag{14}$
and
$\vert N(s) \vert = 1; \tag{15}$
assembling (12) and (13) together we have
$\kappa(s) R \ge 1, \tag{16}$
or
$\kappa(s) \ge \dfrac{1}{R}, \tag{17}$
$OE\Delta$.