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No.

Let $K$ be any infinite lattice with DCC, a bottom element $b$, and no top element. Create $L$ by adding to four elements to $K$: elements $u$ and $v$ that are incomparable to every element of $K$ and to each other, a new bottom element $0$, and a new top element $1$. (So $0 < u < 1$, $0 < v < 1$, and $0<k<1$ for any element of $K$, but no comparabilities between $u, v, k$ for any $k\in K$.) The DCC is used only to guarantee that $L$ is complete.

$L$ has a join-incomplete endomorphism $f\colon L\to L$, which maps every element of $K$ to its bottom element $b$, and maps every other element of $L$ to itself. But if $S\subseteq L$ is infinite, then it must contain infinitely many members of $K$. If $f_0\colon L\to L$ satisfies $f_0(S)=\{0\}$, then $f_0(b)=0$. But this forces $f_0(u)=f_0(u)\vee f_0(b) = f_0(u\vee b) = f_0(1)$ and similarly $f_0(v)=f_0(1)$, hence $0 = f_0(0) = f_0(u\wedge v) = f_0(1)\wedge f_0(1) = f_0(1)$. That is, any endomorphism of $L$ that maps an infinite set to $0$ is constant.

No.

Let $K$ be any infinite lattice with DCC, bottom element $b$, and no top element. Create $L$ by adding to four elements to $K$: elements $u$ and $v$ that are incomparable to every element of $K$ and to each other, a new bottom element $0$, and a new top element $1$. (So $0 < u < 1$, $0 < v < 1$, and $0<k<1$ for any element of $K$, but no comparabilities between $u, v, k$ for any $k\in K$.) The DCC is used only to guarantee that $L$ is complete.

$L$ has a join-incomplete endomorphism $f\colon L\to L$, which maps every element of $K$ to its bottom element $b$, and maps every other element of $L$ to itself. But if $S\subseteq L$ is infinite, then it must contain infinitely many members of $K$. If $f_0\colon L\to L$ satisfies $f_0(S)=\{0\}$, then $f_0(b)=0$. But this forces $f_0(u)=f_0(u)\vee f_0(b) = f_0(u\vee b) = f_0(1)$ and similarly $f_0(v)=f_0(1)$, hence $0 = f_0(0) = f_0(u\wedge v) = f_0(1)\wedge f_0(1) = f_0(1)$. That is, any endomorphism of $L$ that maps an infinite set to $0$ is constant.

No.

Let $K$ be any infinite lattice with DCC, a bottom element $b$, and no top element. Create $L$ by adding to four elements to $K$: elements $u$ and $v$ that are incomparable to every element of $K$ and to each other, a new bottom element $0$, and a new top element $1$. (So $0 < u < 1$, $0 < v < 1$, and $0<k<1$ for any element of $K$.) The DCC is used only to guarantee that $L$ is complete.

$L$ has a join-incomplete endomorphism $f\colon L\to L$, which maps every element of $K$ to its bottom element $b$, and maps every other element of $L$ to itself. But if $S\subseteq L$ is infinite, then it must contain infinitely many members of $K$. If $f_0\colon L\to L$ satisfies $f_0(S)=\{0\}$, then $f_0(b)=0$. But this forces $f_0(u)=f_0(u)\vee f_0(b) = f_0(u\vee b) = f_0(1)$ and similarly $f_0(v)=f_0(1)$, hence $0 = f_0(0) = f_0(u\wedge v) = f_0(1)\wedge f_0(1) = f_0(1)$. That is, any endomorphism of $L$ that maps an infinite set to $0$ is constant.

No.

Let $K$ be any infinite lattice with DCC, a bottom element $b$, and no top element. Create $L$ by adding to four elements to $K$: elements $u$ and $v$ that are incomparable to every element of $K$ and to each other, a new bottom element $0$, and a new top element $1$. (So $0 < u < 1$, $0 < v < 1$, and $0<k<1$ for any element of $K$, but no comparabilities between $u, v, k$ for any $k\in K$.) The DCC is used only to guarantee that $L$ is complete.

$L$ has a join-incomplete endomorphism $f\colon L\to L$, which maps every element of $K$ to its bottom element $b$, and maps every other element of $L$ to itself. But if $S\subseteq L$ is infinite, then it must contain infinitely many members of $K$. If $f_0\colon L\to L$ satisfies $f_0(S)=\{0\}$, then $f_0(b)=0$. But this forces $f_0(u)=f_0(u)\vee f_0(b) = f_0(u\vee b) = f_0(1)$ and similarly $f_0(v)=f_0(1)$, hence $0 = f_0(0) = f_0(u\wedge v) = f_0(1)\wedge f_0(1) = f_0(1)$. That is, any endomorphism of $L$ that maps an infinite set to $0$ is constant.

No.

Let $K$ be the lattice of finitely generated subspaces of an infinite dimensional vector space, with bottom element labelled $b$. Create $L$ by adding to four elements to $K$: elements $u$ and $v$ that are incomparable to every element of $K$ and to each other, a new bottom element $0$, and a new top element $1$. (So $0 < u < 1$, $0 < v < 1$, and $0<k<1$ for any element of $K$.)

The lattice $K$ has a join-incomplete endomorphism $f\colon L\to L$, which maps every element of $K$ to its bottom element $b$, and maps every other element of $L$ to itself. But if $S\subseteq L$ is infinite, then it must contain infinitely many members of $K$. If $f_0\colon L\to L$ satisfies $f_0(S)=\{0\}$, then $f_0(b)=0$. But this forces both $f_0(u)=f_0(u)\vee f_0(b) = f_0(u\vee b) = f_0(1)$ and similarly $f_0(v)=f_0(1)$, hence $0 = f_0(0) = f_0(u\wedge v) = f_0(1)\wedge f_0(1) = f_0(1)$. That is, any endomorphism of $L$ that maps an infinite set to $0$ is constant.

No.

Let $K$ be any infinite lattice with DCC, a bottom element $b$, and no top element. Create $L$ by adding to four elements to $K$: elements $u$ and $v$ that are incomparable to every element of $K$ and to each other, a new bottom element $0$, and a new top element $1$. (So $0 < u < 1$, $0 < v < 1$, and $0<k<1$ for any element of $K$.) The DCC is used only to guarantee that $L$ is complete.

$L$ has a join-incomplete endomorphism $f\colon L\to L$, which maps every element of $K$ to its bottom element $b$, and maps every other element of $L$ to itself. But if $S\subseteq L$ is infinite, then it must contain infinitely many members of $K$. If $f_0\colon L\to L$ satisfies $f_0(S)=\{0\}$, then $f_0(b)=0$. But this forces $f_0(u)=f_0(u)\vee f_0(b) = f_0(u\vee b) = f_0(1)$ and similarly $f_0(v)=f_0(1)$, hence $0 = f_0(0) = f_0(u\wedge v) = f_0(1)\wedge f_0(1) = f_0(1)$. That is, any endomorphism of $L$ that maps an infinite set to $0$ is constant.