 Solution

Yes. Suppose $f$ is not identically zero. Thenis a nonempty open set, and $fg=0$ forces $g=0$ on $U$. Harmonic functions satisfy unique continuation: locally they are real parts of holomorphic functions and hence are real analytic, so a harmonic function vanishing on a nonempty open subset of a connected domain vanishes everywhere. Thus $g\equiv 0$. Interchanging $f$ and $g$ proves that one factor must be identically zero.