What is the difference between solutions of the diffusion equation with an imaginary diffusion coefficent and the wave equation's?

Both Schrödinger and Wave Equation have plane wave solutions, that's right. The difference is the dispersion relation, which is quadratic for the Schrödinger equation and linear for the wave equation. This is important, because the Schrödinger equation was designed to correctly reproduce the quadratic dispersion relation that was observed for electrons.

(You can show that by fourier-transforming both equations both in space and time and solving for $\omega(k)$.)

I don't think your understanding is fundamentally flawed. There are many ways to see the resemblance between the heat/diffusion equation and Schrödinger equation, one of which being the stochastic interpretation mentioned in one of the answers to the question cited as possible duplicate. According to wikipedia, Erwin Schrödinger himself was aware of this as well. Here is a question from mathoverflow that puts this in a more mathematical context.

However, in the most common cases, the correspondence between the common solutions of both equations cannot be easily exploited. This is because PDE problems must be formulated with initial/boundary conditions, and the boundary conditions usually used for heat/diffusion equation and Schrödinger are different. A problem involving the heat equation usually specifies the initial data on all points in space, with Dirichlet or Neumann boundary conditions, and we are interested in the time evolution of the initial data. In the free space Schrödinger equation, we could of course specify an initial wave function and look at its evolution, which would make it equivalent to the previous heat problem. However, more often we see the time dependence separated out, and we want to find out the spacial dependence of the solution given a boundary condition, and usually the only meaningful boundary condition is a homogeneous Dirichlet BC. Moreover, the Schrödinger equation with a nonzero potential is more difficult to relate to the heat problem.

It seems like one can transform the diffusion equation to an equation that can replace the wave equation since the solutions are the same.

I think replacing a real constant to an imaginary constant is deceivingly simple: even though the equations and solutions are written the same, they encode different information. This is because that now you are working in two dimensions.

If you have an equation of the form: $$ \frac{\partial u(x,t)}{\partial t} = iD\frac{\partial ^2u(x,t)}{\partial x^2} $$

Then writing $u(x,t) = q(x,t) +ip(x,t)$ where $q$ and $p$ are real valued, you get $$ \frac{\partial q(x,t)}{\partial t}+i\frac{\partial p(x,t)}{\partial t} = iD\frac{\partial ^2q(x,t)}{\partial x^2} -D\frac{\partial ^2p(x,t)}{\partial x^2} $$ And so a system of coupled differential equations: $$ \frac{\partial q(x,t)}{\partial t} = -D\frac{\partial ^2p(x,t)}{\partial x^2} $$ $$ \frac{\partial p(x,t)}{\partial t} = D\frac{\partial ^2q(x,t)}{\partial x^2} $$

So you see the seemingly innocent change actually brings you to a quite different set of equations.

These equations are also not quite the same as the wave equation, in particular, as already mentioned in another answer, you get a different dispersion relation. Taking a time derivate of one of the equations leads to: $$ \frac{\partial^2 q(x,t)}{\partial t^2} = -D^2\frac{\partial ^4q(x,t)}{\partial x^4} $$ So that the frequency is proportional to the square of the wavenumber.