Limits of (commutative) monoids can simply by constructed with pointwise operations from the underlying limits of sets. This shows immediately that $\mathbf{CMon}$ is complete and that $U$ is continuous. To show the solution set condition, let $M$ be a commutative monoid and let $f : S \to U(M)$ be a map. Consider the submonoid $M'$ of $M$ generated by the image of $f$. Explicitly, $M'$ consists of all $\mathbb{N}$-linear combinations of elements of the form $f(s)$, where $s \in S$. The next step is to bound the size of the underlying set of $M'$. This is usually done with cardinal numbers, but actually one does not need cardinal numbers at all
for this argument. Notice that there is a surjective map*
$\coprod_{n \in \mathbb{N}} (\mathbb{N} \times S)^n \to U(M').$
which maps $((m_1,s_1)),\dotsc,(m_n,s_n))$ to $m_1 f(s_1) + \cdots + m_n f(s_n)$. It follows that $U(M')$ is isomorphic to a subset of the set $T := \coprod_{n \in \mathbb{N}} (\mathbb{N} \times S)^n$ (which only depends on $S$). By structure transport, it follows that $M'$ is isomorphic to a monoid whose underlying set is a subset of $T$. But there is only a set of such monoids. Therefore, a solution set consists of those monoids whose underlying set is a subset of $T$.
By the way, a similar argument shows that for every finitary algebraic category $\mathcal{C}$ the forgetful functor $\mathcal{C} \to \mathbf{Set}$ has a left adjoint. In fact, it is even monadic.
For commutative monoids, though, the simplest way is just a direction construction: the underlying set of the free commutative monoid on $S$ is the set of functions $S \to \mathbb{N}$ with finite support (these formalize linear combinations with $\mathbb{N}$-coefficients in $S$), and the operation comes from $\mathbb{N}$. For monoids and groups (or lie algebras etc.), direct constructions are not so straight forward anymore and the above method gives a really slick existence proof.
*Alternatively, you can use a surjective map $\coprod_{n \in \mathbb{N}} S^n \to U(M')$.
The problem -in short, and quite roughly speaking- is that you need some assumptions ensuring that the colimit/coend to compute the left Kan extension $Lan_F1$ exists, as the domain might be a large category.
Conditions like the solution set condition or other forms of adjoint functor theorem are usually of the form "the colimit you would compute in order for the adjoint to exist can actually be computed on a small category, so it does exist".
Best Answer
The forgetful functor from the category of groups to the category of monoids has a left adjoint, which may be considered a non-commutative generalisation of the Grothendieck group construction. The colimit formula is not very useful for obtaining an explicit description – as you observed, it doesn't use anything special about the categories or functors in question. Here is a concrete construction.
Let $M$ be a monoid, written multiplicatively with unit $1$. We consider the group $G$ defined as follows. The elements of $G$ are finite words of the form $a b^{-1} c d^{-1} \cdots$ where $a, b, c, d, \ldots$ are elements of $M$, subject to the following equivalence relations:
The group operation of $G$ is essentially concatenation: given words $a b^{-1} \cdots$ and $c d^{-1} \cdots$, the product is $a b^{-1} \cdots c d^{-1} \cdots$, with a $1$ or $1^{-1}$ inserted in the middle if necessary. The inverse of $a b^{-1} c d^{-1} \cdots$ is $\cdots d c^{-1} b a^{-1}$. It is straightforward, if tedious, to verify that this is indeed a group such that every monoid homomorphism $M \to H$ factors as the obvious map $M \to G$ followed by a group homomorphism $G \to H$, and that this factorisation is unique.