Note that $\mathcal A$ is really the same thing as $(\mathcal A^{\mathrm{op}})^{\mathrm{op}}$. So, if we apply the second definition to a functor $X : \mathcal A = (\mathcal A^{\mathrm{op}})^{\mathrm{op}} \to \mathsf{Set}$, we get that we call it representable if
$$
X \cong \mathrm{Hom}_{\mathcal A^{\mathrm{op}}}(-, A) \cong \mathrm{Hom}_{\mathcal A}(A, -).
$$
However, I think this perspective is a lot less useful than the perspective in the comments, that these are really just two separate definitions.
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
Hint: If $F$ is left adjoint to $G$ (just assume that it exists for the moment), then $F(*)=X$, and $F$ preserves colimits, in particular coproducts. Now compute $F(S)$ for an arbitrary set $S$. After that, show that, in fact, $F$ defined by ... exists and is left adjoint to $G$.