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Question: Are these adjoint functors to/from the category of monoids with semigroup homomorphisms?
Answer: The well known functor $F: \bf Semigroup \to \bf Monoid$ is the left adjoint to the forgetful functor $G: \bf Monoid \to \bf Semigroup$. To see that $\hom(F(X),Y) \equiv \hom(X,G(Y))$ natural in the variables $X$ and $Y$, explicitly writing down the natural transformations $\varphi$ and $\psi$ inverse to each other helps.


For each semigroup homomorphism $g:X\to G(Y)$, the monoid homomorphism $\psi(g):F(X)\to Y$ maps the adjoint identity element to the identity element, and is otherwise identical to $g$.
For each monoid homomorphism $f:F(X)\to Y$, the semigroup homomorphism $\varphi(f):X \to G(Y)$ is the restriction of $f$ to $X$.


The compositions $\varphi\circ \psi$ and $\psi \circ \varphi$ are both identities.



If we explicitly spell out the corresponding transformations $\varphi$ and $\psi$ for $F_H$ and $G_H$, then $\psi \circ \varphi$ is an identity, but $\varphi\circ \psi$ is only a projection instead of a true identity (semigroup homomorphism get projected to monoid homomorphisms).

If we explicitly spell out the corresponding transformations $\varphi$ and $\psi$ for $F_O$ and $G_O$, then $\varphi\circ \psi$ is an identity, but $\psi \circ \varphi$ is only a projection instead of a true identity (semigroup homomorphism get projected to monoid homomorphisms) when $X$ is not a monoid.



As Martin Brandenburg wrote in a comment, it is not clear whether $F_H$ and $F_O$ are functors at all. In fact, they fail to be functors, because $F(g\circ f)=F(g)\circ F(f)$ fails.

For $F_H$ look at
$\begin{matrix}
&amp; f &amp; &amp; g &amp; \\
&amp; &amp; * &amp; \longrightarrow &amp; * \\
* &amp; \longrightarrow &amp; * &amp; \longrightarrow &amp; * \\
* &amp; \longrightarrow &amp; * &amp; {}^{\underline{\quad}\nearrow}
\end{matrix}$. We get
$\begin{matrix}
&amp; F_H(f) &amp; &amp; F_H(g) &amp; \\
&amp; &amp; * &amp; \longrightarrow &amp; * \\
* &amp; {}^{\underline{\quad}\nearrow} &amp; * &amp; \longrightarrow &amp; * \\
* &amp; \longrightarrow &amp; * &amp; {}^{\underline{\quad}\nearrow}
\end{matrix}$, so
$\begin{matrix}
F_H(g) \circ F_H(f)\\
\begin{matrix}
* &amp; \longrightarrow &amp; * \\
* &amp; \longrightarrow &amp; *
\end{matrix}\end{matrix}$.

But
$\begin{matrix}
&amp; g \circ f &amp; * \\
* &amp; \longrightarrow &amp; * \\
* &amp; {}^{\underline{\quad}\nearrow}
\end{matrix}$, so we get
$\begin{matrix}
F_H(g \circ f)\\
\begin{matrix}
* &amp; \longrightarrow &amp; * \\
* &amp; {}^{\underline{\quad}\nearrow} &amp; *
\end{matrix}\end{matrix}$.
Hence $F_H(g) \circ F_H(f)$ and $F_H(g \circ f)$ are different.

Here, the vertical stars "describe" specific semilattices (commutative idempotent semigroups), where the result of a multiplication is the lower of the two involved stars (i.e. the meet or the greatest lower bound). Hence the identity element is given by the highest star in each case. The arrows "describe" specific homomorphisms.

For $F_O$ look at
$\begin{matrix}
&amp; f &amp;  &amp; g &amp; * \\
{\scriptsize\begin{matrix}* \quad *\\*\\
\end{matrix}} &amp; \longrightarrow &amp; * &amp; \longrightarrow &amp; * 
\end{matrix}$. The leftmost semilattice is missing an identity element, hence $F_O$ will adjoint one. We get
$\begin{matrix}
&amp; F_O(f) &amp;  &amp; F_O(g)\\
* &amp; {}_{\overline{\;\cdot\;}\searrow} &amp;  &amp;  &amp; * \\
{\scriptsize\begin{matrix}* \quad *\\*\\
\end{matrix}} &amp; \longrightarrow &amp; * &amp; \longrightarrow &amp; * 
\end{matrix}$, so 
$\begin{matrix}
F_O(g) \circ F_O(f) \\
\begin{matrix}
* &amp; {}_{\overline{\;\cdot\;}\searrow}  &amp; * \\
{\scriptsize\begin{matrix}* \quad *\\*\\
\end{matrix}} &amp; \longrightarrow &amp; * 
\end{matrix}\end{matrix}$.

But
$\begin{matrix}
 &amp; g \circ f &amp; * \\
{\scriptsize\begin{matrix}* \quad *\\*\\
\end{matrix}} &amp; \longrightarrow &amp; * 
\end{matrix}$, so we get
$\begin{matrix}
F_O(g \circ f) \\
\begin{matrix}
* &amp; \longrightarrow &amp; * \\
{\scriptsize\begin{matrix}* \quad *\\*\\
\end{matrix}} &amp; \longrightarrow &amp; * 
\end{matrix}\end{matrix}$.
Hence $F_O(g) \circ F_O(f)$ and $F_O(g \circ f)$ are different.