I feel that this is true but I'm unable to prove it formally: a monoid $M$ is $\omega$-presentable in the category $M$-$\mathbf{Set}$. This is the category of $(X,\rho)$ where $\rho:M\times X\to X$ satisfies $(a\cdot b)\cdot x=a\cdot (b\cdot x)$ and $1\cdot x=x$. The explicit definition of finite presentability is given in the snippet below:
Definition. An object $K$ of a category $\mathcal{K}$ is called finitely presentable provided that its hom-functor
$$\operatorname{hom}(K, {-}) : \mathcal{K} \to \mathbf{Set}$$
preserves directed colimits.
Explicitly: $K$ is finitely presentable iff for each directed diagram
$$D : (I, \le) \to \mathcal{K},$$
each colimit cocone $D_i \overset{c_i}{\rightarrow} C ~ (= \operatorname{colim} D), i \in I$, and each morphism $f : K \to C$ there exists $i$ such that
(1) $f$ factorizes through $c_i$, i.e., $f = c_i \cdot g ~ (i \in I)$ for some $g : K \to D_i$,
and
(2) the factorization is essentially unique in the sense that if $f = c_i \cdot g = c_i \cdot g'$, then $D(i \rightarrow j) \cdot g = D(i \rightarrow j) \cdot g'$ for some $j \ge i$.
Let me first review a couple basic properties of directed limits of $M$-sets:
Lemma 1: Suppose $x \in \varinjlim D(i)$. Then there exists some $i \in I$ and $x_i \in D(i)$ such that $x = c_i(x_i)$.
Proof: Consider $S := \bigcup_{i\in I} \operatorname{im}(c_i)$. It is straightforward to show that this is a sub-$M$-set of $\varinjlim D(i)$; and by construction, each $c_i$ factors (uniquely) through $S$. It follows that the factors of $c_i$ induce a morphism of $M$-sets $\varinjlim D(i) \to S$ such that composing with the inclusion map of $S$ gives the identity map on $\varinjlim D(i)$. This implies that $S$ is all of $\varinjlim D(i)$. $\square$
Lemma 2: Suppose $x_i \in D(i)$, $y_j \in D(j)$ satisfy $c_i(x_i) = c_j(y_j)$. Then there exists $k \ge i, j$ such that $D(i\to k)(x_i) = D(j\to k)(y_j)$ in $D(k)$.
Proof: This one is a bit trickier. One way to see this is to construct an $M$-set of so-called germs of functions. Here, an element is given by some $i \in I$ and a function $f \in \prod_{j\ge i} D(j)$ -- i.e. $f$ is a function with domain $\{ j \in I \mid j \ge i \}$ such that for each $j\ge i$, we have $f(j) \in D(j)$. However, we need to give a condition that two germs $f \in \prod_{j\ge i} D(j)$ and $f' \in \prod_{j\ge i'} D(j)$ are considered equal if there exists $i'' \ge i, i'$ such that $f(j) = f'(j)$ whenever $j \ge i''$. (The proof that this gives an equivalence relation on "pre-germs" should be straightforward.) We can define an action of $M$ on the set of pre-germs of functions, where $m \cdot f$ has the same domain as $f$ and $(m \cdot f)(j) := m \cdot (f(j))$. (Check that this respects the equivalence relation, and therefore induces an $M$-set structure on the set $G$ of germs of functions.)
Now for each $i$, we can define a morphism of $M$-sets $D(i) \to G$ where $x \in D(i)$ maps to the pre-germ with domain $\{ j \in I \mid j \ge i \}$ and function $j \mapsto D(i\to j)(x)$, then to the corresponding germ. It should be a straightforward exercise to show that this forms a compatible collection of morphisms $D(i) \to G$, and therefore induces a morphism $\varinjlim D(i) \to G$.
Now, suppose we have $x_i \in D(i), y_{i'} \in D(j)$ such that whenever $j \ge i, i'$, we have $D(i\to j)(x_i) \ne D(i'\to j)(y_{i'})$. This would imply that the germs $j \mapsto D(i\to j)(x_i)$ and $j \mapsto D(i'\to j)(y_{i'})$ generated by $x_i$ and $y_{i'}$ respectively are unequal in $G$. Therefore, we must have $c_i(x_i) \ne c_{i'}(y_{i'})$ in $\varinjlim D(i)$. $\square$
Note that an alternative proof for both lemmas would be to exploit the usual construction of a directed colimit of $M$-sets: start with $\bigsqcup_{i\in I} (\{ i \} \times D(i))$. Then quotient by the equivalence relation where $(i, x_i) \sim (j, y_j)$ if there is some $k \ge i, j$ with $D(i\to k)(x_i) = D(j\to k)(y_j)$ in $D(k)$. Show that this forms an $M$-set with action of $M$ induced by the action $m \cdot (i, x_i) := (i, m \cdot x_i)$. Show that the canonical morphisms $D(i)$ to the quotient form a compatible collection, and show that this system satisfies the required universal property of the directed colimit. Now, note that under this construction, both lemmas are pretty much immediate.
Now, finally, to show directly that the explicit conditions are satisfied:
(1) Suppose we have a morphism of $M$-sets $f : M \to \varinjlim D(i)$. Then by lemma 1, there exists some $i \in I$ and $x_i \in D(i)$ such that $f(e) = c_i(x_i)$. Now define $g : M \to D(i)$ by $g(m) := m \cdot x_i$. It is straightforward to check that $g$ is a morphism of $M$-sets. Also, for any $m \in M$, $$f(m) = f(m\cdot e) = m \cdot f(e) = m \cdot c_i(x_i) = c_i(m\cdot x_i) = c_i(g(m)),$$ so $f = c_i \circ g$ as required.
(2) Suppose we have two morphisms of $M$-sets $g, g' : M \to D(i)$ such that $c_i \circ g = c_i \circ g'$. Then by lemma 2, there exists some $j\ge i$ such that $D(i\to j)(g(e)) = D(i\to j)(g'(e))$. This implies that for any $m\in M$, $$D(i\to j)(g(m)) = D(i\to j)(g(m\cdot e)) = m \cdot D(i\to j)(g(e)) = m \cdot D(i\to j)(g'(e)) \\ = D(i\to j)(g'(m\cdot e)) = D(i\to j)(g'(m)),$$ so $D(i\to j) \circ g = D(i\to j) \circ g'$ as required.