Let $G$ be a group and $G$-Set the category of left $G$-sets, i.e., sets equipped $X$ with a left action of $G$ on $X$. The forgetful functor from $G$-Set to Set has a right adjoint, which sends each set $S$ to the set of all functions $f:G\to S$, equipped with the $G$-action that sends $(g,f)$ (where $g\in G$ and $f:G\to S$) to the function $(gf):G\to S$ sending any $x\in G$ to $f(xg)$.
Let's say that a monoidal subcategory $Z$ of a monoidal category $(X,\otimes,I)$ is one for which
- Morphisms $f,g\in Z$ imply $f\otimes g\in Z$
- Objects $A,B,C\in Z$ imply the associators $(A\otimes B)\otimes C\to A\otimes(B\otimes C)$ are in $Z$
- $I\in Z$
- An object $A\in Z$ implies the unitors $I\otimes A\cong A\cong A\otimes I$ are in $Z$.
Evidently the intersection of families of monoidal subcategories is a monoidal subcategory. Therefore, given a functor $F\colon Y\to X$ there is a smallest monoidal subcategory $MF(Y)$ containing the image of $F(Y)$. In particular, $F\colon Y\to X$ factors as $Y\to MF(Y)\hookrightarrow X$. Moreover, $MF(Y)\hookrightarrow X$ is a monoidal functor, i.e. a functor in the category of small monoidal categories.
I now claim the morphisms in this subcategory are exactly the composites of morphisms of the form $F(f_1)\otimes F(f_2)\otimes...\otimes F(f_n)$ (wih various parenthesizations) for $f_i$ morphisms in $X$, and appropriate unitors and associators.
It follows that $MF(Y)$ has cardinality bounded by $\kappa_Y$, where $\kappa_Y$ is the smallest infinite cardinal bounding the cardinality of $Y$.
If $\lambda$ is its cardinal, then the isomorphism $\lambda\cong UM(Y)$ of $\lambda$ with the set of morphisms of $MF(Y)$ induces a monoidal category structure on $\lambda$ so that the resulting monoidal category is isomorphic to $MF(Y)$.
Thus the functor $F\colon Y\to X$ factors as $Y\to M\to X$ where $M\to X$ is a monoidal functor, and $UM$ is a cardinal bounded by $\kappa_Y$. Since the set of cardinals bounded by $\kappa_Y$ is a set, and since each set has a set of monoidal structures, and since between any two categories there is a set of functors between them, it follows that for every category $Y$ there is only a set of functors $Y\to M$ where $M$ is a monoidal category with $UM$ a cardinal bounded by $\kappa_Y$.
By the previous discussion, this is a solution set for the forgetful functor from small monoidal categories to small categories: any functor $F\colon Y\to X$ factors as $Y\to M\to X$ where $M\to X$ is a monoidal functor with $UM$ a cardinal bounded by $\kappa_Y$.
Best Answer
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 $\coprod_{n \in \mathbb{N}} S^n \to U(M')$.