Let me just point out that the equivalence claimed in Simon Henry's answer is not correct.

There is a unique map of sets $\{1\} \to \{1\}$, but there are two cogroup homomorphisms $F(\{1\}) \to F(\{1\})$, namely $x \mapsto x$ and $x \mapsto 1$. Even more trivially, there is *no* map $\{1\} \to \emptyset$, but there is (exactly) one cogroup homomorphism $F(\{1\}) \to F(\emptyset)$.

This is a good lesson that a category is more than just its objects.

The category of cogroups in $\mathbf{Grp}$ is equivalent to $\mathbf{Set}_*$. The equivalence maps a pointed set $(X,x_0)$ to the free group on $X$ modulo $\langle \langle x_0 \rangle \rangle$ (which is isomorphic to the free group on $X \setminus \{x_0\}$, but with this POV the functoriality is less clear). This is explained (in a much larger framework) in Bergman's *Invitation to general algebra* in Section 10.7.

So it remains to recover $\mathbf{Set}$ from $\mathbf{Set}_*$. I will add to this post when I find something.