Why is the forgetful functor representable?

If representable meant $F$ is equal to $\mathrm{Hom}(R, -)$, then, as you can see from your (correct) observations, the notion would be of practically no use. Representability means instead that the functor is naturally isomorphic to $\mathrm{Hom}(R, -)$ for some $R$.

Loosely speaking, it is a general principle of Category theory that you shouldn't ask whether two things are equal, but only whether they're isomorphic instead. In particular your monoids $A$ and $B$ are isomorphic, which means that the sets $\mathrm{Hom}(ℕ, A)$ and $\mathrm{Hom}(ℕ, B)$ are isomorphic as well, so there's no contradiction with representability of $U$.

When you have an adjunction $\mathsf{F} \dashv \mathsf{U}$ $$\mathsf{F}: \mathsf{Set} \leftrightarrows \mathsf{K}: \mathsf{U},$$ where $\mathsf{U}$ is a forgetful like functor, $\mathsf{U}$ is always representable. This is due to a very specific entanglement that is characteristic of the category of sets (and in general will kinda apply for $\mathcal{V}$ in $\mathcal{V}$-$\mathsf{Cat}$ when $\mathcal{V}$ is monoidal closed). In fact $$\mathsf{U}(\_) \cong \mathsf{Set}(1, \mathsf{U}(\_)) \cong \mathsf{K}(\mathsf{F}1, (\_)),$$ and thus the forgetful functor is representable and is represented by the free algebra over the one element set. This is evident when $\mathsf{K}$ is $\mathsf{R}$-$\mathsf{Mod}$, groups, monoids and algebraic structures in general.

Notice that in the proof I used isomorphic (!!!) functors!