A morphism \$$f:F(m)\rightarrow n\$$ is an element \$$f \in N\$$ such that \$$f\circ_N F(m) = n\$$.
Likewise, a morphism \$$g:m\rightarrow G(n)\$$ is an element \$$g \in M\$$ such that \$$g\circ_M m = G(n)\$$.

Thus we have \$$g \circ_M m = G(n) = G(f \circ_N F(m)) = G(f) \circ_M G(F(m))\$$. If we consider the case \$$m = 1_M\$$ and recall that \$$G(F(1_M))=G(1_N)=1_M\$$, this becomes \$$g=G(f)\$$ so we have the mapping in one direction.

In the other direction, using \$$g=G(f)\$$ and \$$n = 1_N\$$, I get the condition \$$g \circ_M G(F(m)) = g \circ_M m = 1_M\$$, i.e., for any \$$m \in M\$$ we must have \$$g \circ_M m = 1_M\$$ if and only if \$$g \circ_M G(F(m)) \$$. So for all elements \$$m \in M\$$ that have a left-inverse \$$g\$$, this must also be the left-inverse of \$$G(F(m))\$$. This would surely work if \$$G\$$ was a left-inverse of \$$F\$$, but it is not clear to me that it is necessary - nor what monoid concept this corresponds to.