Klein-Gordon Propagator for spatial separation $x - y = (0, r)$

The easiest way to proceed is to notice that the first integral can only depend on $r := |vec{r}|$. To convince yourself of this, you can rotate $vec{r}$ by any rotation $R in SO(3)$, and you notice that you can always "unwind" this rotation by a redefinition of the momentum variable being integrated (if you take $vec{r}to Rvec{r}$ then change variables to $R^{T} vec{p} := vec{p}'$, where the Jacobian is just $1$). Since this works for any $R$ you know this is only a function of $r$.

Because of this fact, you can make your life easier by picking $vec{r} = ( 0, 0, r )$ (notice that $|( 0, 0, r )| = r$ is still true). If you plug this into the first integral you get $$ int frac{d^3 vec{p}}{(2pi)^3 } frac{1}{2 E_p} e^{- i vec{p} cdot vec{r}} = int frac{d^3 vec{p}}{(2pi)^3 } frac{1}{2 E_p} e^{- i p_3 r} $$ in spherical coordinates this becomes $$ cdots = int frac{dp ; dtheta ; dphi}{(2pi)^3 } frac{p^2 sintheta}{2 E_p} e^{- i r p costheta } . $$ I leave it up to you to simplify this to your expression ($phi$ integrates easily, and use the coordinate change $mu = costheta$).