The answer is no.

I will refer to these lattices (the labels are not meaningful - I took the images from a paper). Figure 1 depicts a lattice $A$ with the following properties:

- $A$ is meet semidistributive.
- Every injective endomorphism of $A$ is join-complete.
- If $\theta$ is any nonzero congruence on $A$, then $A/\theta$ is finite. Moreover, the natural map $A\to A/\theta$ is either join-complete or else the image $A/\theta$ is not meet semidistributive.
- The lattice $B$ in Figure 2 is a homomorphic image of $A$ (map $\overline{\overline{a}}$ to $x$, map each $b_{2n}$ to $y$, each $c_{2n+1}$ to $z$).

So take $L$ to be the ordinal sum of $A$ and $B$ ($B$ on top of $A$). Call the least and largest elements $0$ and $1$. This lattice is complete.

Define $f\colon L\to L$ by taking the (bottom) copy of $A$ in $L$ to the (top) copy of $B$, according to the homomorphism described in 4 above, and taking the top copy of $B$ to $1$. $f$ is a join-incomplete, since $f$ does not preserve the join of the central $\omega$-chain. (In fact, $f$ does not preserve the join of any $\omega$-chain.)

**Claim.** $L$ has no contracting join-incomplete endomorphism.

*Sketch.*
Assume that $g\colon L\to L$ is a join-incomplete endomorphism. $g$ must be join-incomplete restricted to $A$, since $A$ contains all nonprincipal ideals of $L$ and their joins. Hence, by 2 and 3 above, $g$ is noninjective on $A$, and $g(A)$ is not meet semidistributive. This clearly implies $g(A)\not\subseteq A$, since $A$ is meet semidistributive, but in fact you can convince yourself of the stronger fact that $g(A)\subseteq B$. This shows that $g$ is not contracting, since all nonprincipal ideals lie in $A$ while their $g$-images lie in $B$. \\