Here's a poset, call it $P.$
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Define a function $S : P \rightarrow P$ as follows.
$$S(0) = 1, \qquad S(1) = 2, \qquad S(2) = 2$$
Clearly, $S$ is monotone. Now consider the following subposets of $P$, all of which are wide but not full:
Observe that $S$ restricts to a monotone self-map on both $A$ and $B$, but not on $C$.
Question. Under what conditions can we expect a monotone mapping $f : P^n \rightarrow P$ to restrict to a monotone mapping $S^n \rightarrow S$, where $S$ is a wide-but-not-full subposet of $P$?
I'm most interested in the case where $P$ is a meet-semilattice, $S$ is a wide-but-not-full subposet of $P$ that isn't a sub-meet-semilattice, and $f$ is the meet operation $\wedge : P^2 \rightarrow P$. I haven't had much success understanding even the basics of this question; I think the meet restricts to a monotone mapping on each of $A,B$ and $C$ above, and I haven't been able to cook up a situation where it clearly doesn't restrict to a monotone mapping.
For a unary function $S: P\to P$, the only wide subposets of $P$ in which $S$ is still monotone are those made with following rule: if you remove an arrow from $P$, you must remove the corresponding arrow(s) from $S^{-1}(P)$. In your example, the preimage of the arrow $1 \to 2$ via $S$ is the arrow $0 \to 1$. So when you remove $1 \to 2$, $S$ remains montone if and only if you remove $0 \to 1$. That's why $S$ is monotone on $A$ but not on $C$. Since nothing maps to $0$ via $S$, that explains why you can remove the arrow $0 \to 1$ without removing anything else.
Note that you may need to remove more than one arrow as a consequence of the removal of a given arrow to preserve monotonicity. For example, with the order $0 < 1 < 2$ and $0' < 1$ (but $0$ and $0'$ incomparable), a monotone function $S'$ can map both $0$ and $0'$ to $1$ and $1 \mapsto 2, 2\mapsto 2$. In this example, when you remove the arrow $1 \to 2$, you have to remove both $0\to 1$ and $0'\to 1$ to preserve monotonicity of $S'$. The preimage of $1\to 2$ via $S'$ consists of both these arrows.
I have yet to think about the $n$-ary function case in full generality, but the above is enough to construct a counter-example for the semilattice & meet case. I assume you consider an n-ary function monotone by ordering its tuple argument such that $(a_1, \ldots, a_n) \le (b_1, \ldots, b_n)$ iff $a_i \le b_i, \forall i \in \{1,\ldots,n\}$.
Now consider the "rhombus" [semil]attice given by $0<1<2,\; 0<1'<2$. The monotonic unary function is going to be the meet with the fixed element $1$, i.e. $f(x)=1\wedge x$; it maps the arrow $1'\to 2$ to the arrow $0 \to 1$, so if we remove only the arrow $0 \to 1$ from this [semi]lattice, then $f$ is not monotone on the resulting poset $1<2,\; 0<1'<2$ (which is not a meet-semilattice) because $f(1')=0$ and $f(2) = 1$, but $0$ and $1$ are incomparable in the discompleted rhombus.