Yes. Consider a point on the boundary contained both in the closure of $U$ and $V$ and do a Mobius transform transforming our disk into upper half-plane $\mathbb{H}$ and putting this point to $0$.
Now take the map $(x, y) \mapsto x+y$.
It is open, and its image (from $U \times V$) contains arbitrarily small translations of $U$ and $V$, hence contains $U$ and $V$, hence is $\mathbb{H}$.
Edit: actually, for this proof to work one needs to find such a point on the boundary. It automatically follows if you assume that $U$ and $V$ are simply-connected.