Probably missing something very obvious, sorry if this is a stupid question.
I have to show the function $\psi\left(x, y\right) = \frac{1}{\sqrt{2\pi\mathrm{i}t}} \int_{-\infty}^{\infty}\mathrm{e}^{\mathrm{i}\left(x - y\right)^{\, 2}/4t}\, \psi_{0}\left(y\right)\,\mathrm{d}y$ is a solution of the Schrodinger equation for a free particle, i.e. is a solution of $i\hbar \frac{\partial \psi(x,t)}{\partial t} = \frac{-\hbar^2}{2m} \frac{\partial^2 \psi(x,t)}{\partial x^2}$.
Working through I get that $i\hbar \frac{\partial \psi(x,t)}{\partial t} = -\hbar \frac{\partial^2 \psi(x,t)}{\partial x^2}$. That is, I can't get a $1/2m$ term no matter what I do, and my lecture notes don't contain any relations of $\hbar$ and $m$ so I'm not sure how to finish the proof.
I read in another question on free particles we can "let $\hbar^2/2m = 1$ without loss of generality but I can't find a source on this claim.
Any help or hints appreciated, thanks.
That function certainly can't solve that equation, since the parameters $\hbar$ and $m$ don't occur in it.
It's quite usual to set various constants or combinations thereof to $1$ in physics to simplify expressions. You can either do this simply by choosing units such that the constant expression takes the value $1$, or you can explicitly rescale time and space. See natural units.
Note that in the present case the choice $\frac{\hbar^2}{2m}=1$ that you mention would be unhelpful, as it would leave a factor of $\hbar$ on the left-hand side – what you need is $\frac\hbar{2m}=1$.