Edit: actually, proof works in the assumption that $U$ and $V$ are simply-connected. I do not know whether the answer is "yes" for the covering by an annulus and a concentric circle.

(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 $\mathbb{H}$.