This was asked on a stats test I just took, and I'm convinced it's a typo and should have been $X+Y$, which has a well-known answer. Here is my best guess at the PMF of $Z=XY$ $$ P(Z=z) = P(XY=z) = P(X=\frac{z}{y})P(Y=y) \forall y \text{ s.t. } y\mid z $$ $$ P(Z=z) = e^{-\lambda-\mu} \sum_{i=1}^z \left\{ \begin{array}{ll} \frac{\lambda^{\frac{z}{i}}\mu^i}{\frac{z}{i}!i!} & \text{if} \quad i|z. \\ 0 & \text{else} \end{array} \right. $$
Unless I'm missing something huge, I don't think there's a way to simplify this further into a nice equation, is there?