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
To every differentiable symmetry of the Action of a system, there corresponds a conserved current.
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.
