I disagree with the accepted answer. You do not need a Lipschitz bound on $a$. I assume for simplicity that $a\in L^\infty(\Omega)$.

Step 1. The solution $u$ exists and is unique in $H^1_0(\Omega)$, by Lax-Milgram for example.

Step 2. Write $g:=f-au$. Then $u$ satisfies $-\Delta u = g$ in $\Omega$, so provided $\Omega$ is $C^1$ for example, $u\in H^2(\Omega)\cap H^1_0(\Omega)$. So

2.a If $n$ the dimension of the ambient space verifies $n<4$ then $u\in C^{0,1/4}(\Omega)$ and therefore
$$
au \in H^{1}(\Omega) \mbox{ when } a\in H^1(\Omega).
$$
2.b If $n>4$ then $u\in L^{\frac{2n}{n-4}}(\Omega)$ and therefore
$$
au \in H^{1}(\Omega) \mbox{ when } a\in W^{1,n/2}(\Omega).
$$
2.c If $n=4$, $a\in W^{1,2+\epsilon}(\Omega)$ with $\epsilon>0$ is enough.

3.a. Suppose $n \geq 6$, and $a\in H^{1}\cap L^\infty$. If we allow $f\in H^1$, as suggested in the question, then $f\in L^{\frac{2n}{n-2}}$, $au \in L^{\frac{2n}{n-4}}$ so $u \in W^{2,\frac{2n}{n-4}}$ and as $\frac{2}{n-4}< 2$, $u$ is Holder, therefore bounded and we are back to 2.a

3.b and 3.c, n=5 and n=4, argue as in 3.a but it bootstrap twice instead of once.

Therefore in all dimension, $a\in H^{1}\cap L^\infty$ is enough.