As a simple matter of a general rule, you cannot derive a more fundamental theory from a less fundamental theory using any form of deductive logic. So no, you cannot possibly derive the Schrödinger equation from the classical wave equation. If you see anyone claiming to have done such witchcraft, beware of them.
So, the question remains (and always was) as to whether the resemblance between $y=A\sin(kx-wt)$ and $\Psi=Ae^{i(kx-wt)}$ is purely accidental or it can be understood more closely. Now, first of all, this question doesn't translate to the question you ask in the title of your question. Because, $y=A\sin(kx-wt)$ is not the classical wave equation and neither is $\Psi=Ae^{i(kx-wt)}$ the Schrödinger equation. They are specific solutions to the respective equations.
In particular, $\Psi=Ae^{i(kx-wt)}$ is a solution to the free particle Schrödinger equation. The Schrdödinger equation, in the position basis, reads $i\hbar\frac{\partial \Psi(x,t)}{\partial t}=-\frac{\hbar^2}{2m}\frac{\partial ^2\Psi(x,t)}{\partial x^2}+V(x)\Psi(x,t)$. I will omit the discussion about the time-independence of the potential $V(x)$ and the caveats of such a claim, because, the case we are interested in is that of a free particle where the potential simply vanishes for all positions and for all time. So, the free-particle Schrödinger equation reads $\frac{\partial \Psi(x,t)}{\partial t}=i\frac{\hbar}{2m}\frac{\partial ^2\Psi(x,t)}{\partial x^2}$. This is a diffusion equation (not a wave equation) with a purely imaginary diffusion coefficient. Such an equation admits plane-wave solutions of the form $\Psi=Ae^{i(kx-wt)}+Be^{i(-kx-wt)}$ where the dispersion relation reads $w(k)=\frac{\hbar k^2}{2m}$, Here, $k$ is related to energy as $E=\frac{\hbar^2k^2}{2m}$. The $e^{i(kx-wt)}$ part of the solution has a definite momentum $p=\hbar k$ and the $e^{i(-kx-wt)}$ has a definite momentum $p=-\hbar k$. So, a plane-wave solution with a specific energy $E$ and a specific momentum $p=\hbar k$ is simply $\Psi=Ae^{i(kx-wt)}$--the solution that you are interested in.
On the other hand, $y=A\sin(kx-wt)$ is a solution to the classical wave equation $\frac{\partial^2y(x,t)}{\partial t^2}=c^2\frac{\partial ^2y(x,t)}{\partial x^2}$ where $c$ is a positive constant. This equation admits plane-wave solutions of the form $y(x,t)=Ae^{i(kx-wt)}+Be^{i(kx+wt)}$ with the dispersion relation $w(k)=|k|c$. This is a solution with a specific wave-number $k$ and thus, a specific angular frequency $w(k)$. The $e^{i(kx-wt)}$ part of the solution has the velocity of propagation $\frac{kc}{|k|}$ and the $e^{i(kx+wt)}$ part has the velocity of propagation $-\frac{kc}{|k|}$. So, a plane-wave solution with a specifc wave-number $k$ and a specific direction of propagation $\frac{k}{|k|}$ is simply $y(x,t)=Ae^i(kx-wt)$--the purely imaginary part of which is $A\sin(kx-wt)$--the solution that you are interested in.
So, the resemblance that we set out to explore arises out of the fact that the free-particle Schrödinger equation, which is a diffusion equation with a purely imaginary diffusion constant, admits plane-wave solutions just like a classical wave equation does. There is, however, a crucial difference between the two seemingly resembling solutions: their dispersion relation. In particular, the free-particle solution of the Schrödinger equation has a quadratic dispersion relation while the solution of the classical wave-equation has a linear dispersion relation.
So, despite both of them being wave solutions, we can see that they are structurally quite different as the structure of a wave-solution is contained in its dispersion relation.
Related: What is the difference between solutions of the diffusion equation with an imaginary diffusion coefficent and the wave equation's?
