It suffices to show this for l.o. free monoids on finite alphabets, as follows from the Compactness Theorem in logic - cf. any text on First Order Logic. [This principle can be applied to a range of similar problems.]
Indeed, let $\mathbb M = (M, \cdot, \le)$ be any linearly ordered free monoid, and consider the first-order language $\mathcal{L} = (\circ, \preceq, \{x_m|m \in M\})$ consisting of a binary function symbol $\circ$ to represent multiplication, a binary relation symbol $\preceq$ representing ordering, plus an individual constant symbol $x_m$ for each element $m$ of the set $M$.
Then let $T$ be the $\mathcal{L}$-theory having the following axioms:
- the usual axioms for linearly ordered groups, expressed using $\circ$ and $\preceq$
- $x_m \ne x_n$ for all $m, n\in M$ with $m\ne n$ [these axioms ensure that $M$ naturally injects as a subset of any model of $T$ via the interpretation of the $x_m$ constants]
- $x_m \preceq x_n$ for all $m, n\in M$ with $m\le n$ [these make this injection an embedding of l.o. sets]
- $x_m \circ x_n = x_{m\cdot n}$ for all $m, n\in M$ [which make the embedding a monoid homomorphism].
So any model $\mathbb G = (G, \cdot, \le, \{g_m|m \in M\})$ of $T$ will provide the desired l.o. group, with $m\mapsto g_m$ (the interpretation of the constant $x_m$ in $\mathbb G$) giving the desired embedding $\mathbb M\to \mathbb G$.
By the Compactness Theorem, the theory $T$ admits a model iff every finite subset $\Delta$ of (the axioms of) $T$ does. Now only finitely many symbols $x_m$ can occur in sentences belonging to such a $\Delta$, and the finitely many elements $m$ involved generate a l.o. submonoid $\mathbb S$ of $\mathbb M$ which is necessarily free on a finite alphabet. Hence it is enough to consider finitely generated free l.o. monoids.