No.
Let $L$ be the lattice obtained from $\omega+1$ (in its usual order) by adding new elements $a$ and $b$, which are taken to be incomparable to each other and to every element of $\omega+1$ except the bottom, $0$, and the top, $\omega$. Let $f\colon L\to L$ be the map defined by sending some proper tail end of $\omega$ to its first element and every other element of $L$ to itself. (E.g., send every element in the set $\{100, 101, 102, \ldots\}$ to $100$). This map is a lattice homomorphism that is join-incomplete, but the only element of $L$ that maps to $0$ is $0$.