The monoidal center $Z(C)$ of a monoidal category $C$ comes with a forgetful functor $F:Z(C)\rightarrow C$ defined $Z(X,\phi)=X.$ Does $F$ always admit a left adjoint? This is known (Section 3.2.) if $C$ is a tensor category. What about the general case? Tried the General Adjoint Functor Theorem but I'm not even sure if $F$ preserves limits in general.
Category Theory – Forgetful Functor $Z(C)\rightarrow C$ and Its Left Adjoint
adjoint-functorscategory-theorylimitsmonoidal-categoriestensor-products
Related Solutions
If you are just interested in finding some solution set, then there is an easier answer. Arguably this is cheating though, because it uses the description of the left adjoint $F$ of the forgetful functor $U: \mathbf{Gr} \to \mathbf{Set}$.
That being said, let's fix some set $X$. Then I claim that $\{F(X)\}$, where $F$ is the free group on $X$, is a solution set. Indeed, let $f: X \to U(G)$ be any function of sets. Then because $F(X)$ is the free group on $X$ we can extend $f$ to a group homomorphism $\bar{f}: F(X) \to G$ such that $\bar{f} i = f$, where $i: X \to F(X)$ is the obvious injection of the generators.
As I said, this is cheating because it uses a description of the left adjoint of $U$. Indeed, the free group functor $F: \mathbf{Set} \to \mathbf{Gr}$ is the left adjoint of $U$. So the above is just a specific instance of the following fact.
Let $L: \mathcal{C} \rightleftarrows \mathcal{D}: R$ and let $C$ be an object in $\mathcal{C}$. Then $\{ L(C) \}$ is a solution set for $R$, associated to $C$.
Proving this would be a nice exercise. Hint: use the unit $\eta_C: C \to RL(C)$.
The class $S_X$ you proposed is of course going to capture the free group as well, so it will definitely contain enough. As already mentioned in the comments, $S_X$ is technically a proper class. However, this is no issue, because it will be essentially small. That means that there is a set $S_X'$ such that for every group $G \in S_X$ there is $G' \in S_X'$ with $G' \cong G$. So technically $S_X'$ would be our solution set, but we might as well work with $S_X$ and suppress the isomorphisms in our notation (so this is purely for convenience).
To prove that this class is essentially small, you quote another answer where they give a cardinality argument. Basically what they say is that there is some cardinal $\kappa$ such that for all $G \in S_X$ we have $|U(G)| \leq \kappa$. If we fix a set $Y$ then there is only a set of group structures on $Y$. So for every cardinal $\lambda \leq \kappa$ there is, up to isomorphism, only a set of group structures where the underlying set has cardinality $\lambda$. Hence there is, up to isomorphism, only a set of groups where the underlying set has cardinality $\leq \kappa$.
In both answers you quoted it is explained how this cardinality bound is found (although I think in the first one it should be $\mathbb{Z}$ in place of $\mathbb{N}$, but this is of no real consequence).
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 $\coprod_{n \in \mathbb{N}} S^n \to U(M')$.
Best Answer
Consider a nontrivial monoid $M$ with trivial center (in the sense of monoid theory, i.e. the only element of $M$ commuting with all elements of $M$ is $1\in M$). We can consider $M$ as a monoidal category $(\mathcal{M},\otimes)$ in which $\mathcal{M}$ is the discrete category on the elements of $M$, and $\otimes$ is the monoid operation of $M$. Now, since $\mathcal{M}$ only has identity morphisms and $M$ has a trivial center, the center of $(\mathcal{M},\otimes)$ is the trivial category $*$, so $F\colon *\to\mathcal{M}$ is the inclusion of the object $1$. This is a right adjoint functor iff $1$ is a terminal object of $\mathcal{M}$. Since $\mathcal{M}$ has more than one element and is discrete, this cannot be.