Well, I know that if $f$ is a homomorphism from the group $(X,\oplus, x_0)$ to the group $(Y,\otimes, y_1)$ then $f(x_0)$ is $y_1$ and moreover the image of an invertible element of $X$ is an invertible element of $Y$ and the equality $$ f(x^{-1})=f(x)^{-1} $$ holds. So to proving this it is used the cancellation law which is not valid for monoids in general so that I ask to when this result can be true for monoids and it seem to me that it is true if $f$ is an epimorphism, that is a surjective homomorphism. Indeed, if $f$ is surjective then for any $y\in Y$ there exists $x\in X$ such that $$ y=f(x) $$ so that the equality $$ y\otimes f(x_0)=f(x)\otimes f(x_0)=f(x\oplus x_0)=f(x)=y=f(x)=f(x_0\oplus x)=f(x_0)\otimes f(x)=f(x_0)\otimes y $$ holds so that by the uniqueness of the neutral element we conclude that $$ y_0=f(x_0) $$ Moreover, we observe that if $x\in X$ is invertible then $$ f(x)\otimes f(x^{-1})=f(x\oplus x^{-1})=f(x_0)=y_0=f(x_0)=f(x^{-1}\oplus x)=f(x^{-1})\otimes f(x) $$ which prove that $f(x)$ is invertible and moreover by the uniqueness of the inverse the equality $$ f(x^{-1})=f(x)^{-1} $$ holds.
Moreover, we conclude that if $f$ is just an isomorphism then any $x$ in $X$ is invertible if and only if $f(x)$ is invertible in $Y$: indeed, if $f$ is an isomorphism then it is a epimorphism so that if $x$ is invertible then what above ensures that $f(x)$ is invertible; vice versa, the invers of an isomorphism is an isomorphism and so a empimorphism so that if $f(x)$ is invertibile then what above implies that $f^{-1}\big(f(x)\big)=x$ is invertible.
So I like to know if actually what stated about monoids is true and so if it is well proved: could someone help me, please?
Claim: Given a surjective semigroup homomorphism $f$ between two monoids $(X,\oplus,x_0),(Y,\otimes,y_1)$ (This means $f(x\oplus x') = f(x)\otimes f(x') $) then it is already a monoid homomorphism (additionally $f(x_0)=y_1$).
Proof: As $f$ is surjective for any $y\in Y$ there exists $x\in X$ such that $f(x) = y$. Now $$f(x_0)\otimes y = f(x_0)\otimes f(x) = f(x_0\oplus x) = f(x) = y$$ and $$y\otimes f(x_0) = f(x)\otimes f(x_0) = f(x\oplus x_0) = f(x) = y$$ By uniqueness of the unit element $f(x_0) = y_1$. $\square$
Note that we need the surjectivity so that we can do the calculation for all $y \in Y$ or else the statement is not true: see here for details.
Now for any monoid homomorphism $f(x)$ is clearly invertible if $x$ is invertible as it has the inverse $f(x^{-1})$:
$$f(x)\otimes f(x^{-1}) = f(x\oplus x^{-1}) = f(x_0) = y_1 = f(x_0) = f(x^{-1}\oplus x) = f(x^{-1})\otimes f(x)$$
The uniqueness of the unit element in a semigroup $(M,\times)$ is not hard either, assume both $e$ and $f$ fulfill the criteria, then: $$e = e \times f = f$$