In both my textbook (Hungerford's Algebra), and in class, it is claimed that Monoid Homomorphisms are not required to preserve the identity. Interestingly enough, the Wikipedia page for Monoids requires Monoid Homomorphisms to preserve the identity element: https://en.wikipedia.org/wiki/Monoid#Monoid_homomorphisms. I haven't found an example of the former, so I thought I'd prove the opposite statement.
I believe that I have proved the opposite assertion, based on the proof that I used to show that Group Homomorphisms preserve the identity. Since I don't use any information stating that elements are invertible, I think my proof is still valid.
Let $M, N$ be monoids, and let $f:M\rightarrow N $ be a homomorphism of monoids. Let $m,e_{M} \in M$ be an arbitrary element and the identity in $M$ respectively .
Then: $$f(m) = f(m\cdot e_{M}) = f(m)\cdot f(e_{M})$$ $$f(m) = f(e_{M} \cdot m) = f(e_{M}) \cdot f(m)$$ Thus: $$f(m)\cdot f(e_{M}) = f(e_{M}) \cdot f(m) = f(m), \forall m \in M $$ This seems to imply my assertion. Is there anything wrong with my proof?