Your edited question is:

One way to explicitly describe the free abelian group on a set S is as the direct sum of S copies of . Is there a categorical description of this construction, i.e. one which doesn't refer to elements, only morphisms?

That description is **already** categorical. The direct sum of abelian groups is the coproduct, so once you are convinced that the free group on a single element exists (i.e. $\mathbb{Z}$) then you have them all. This is implicit in Theo's answer, but I guess he was replying to your original question. More generally, in a variety of algebras (in the sense of universal algebra), once you have the free *thing* on one generator then you have all of them. The universal property says that we are looking for something such that

$$
\operatorname{Hom}_{\operatorname{Set}}(S,|V|) \cong \operatorname{Hom}_{\operatorname{Alg}}(F(S),V)
$$

(where $\operatorname{Alg}$ stands for our variety of algebras) but since $S = \coprod_S \lbrace \star \rbrace$, we already have

$$
\begin{aligned}
\operatorname{Hom}_{\operatorname{Set}}(S,|V|) &\cong \operatorname{Hom}_{\operatorname{Set}}(\coprod_S \lbrace \star \rbrace,|V|) \\
&\cong \operatorname{Hom}_{\operatorname{Set}}(\lbrace \star \rbrace,|V|)^S \\
&\cong \operatorname{Hom}_{\operatorname{Alg}}(F(\lbrace \star \rbrace),|V|)^S \\
&\cong \operatorname{Hom}_{\operatorname{Alg}}(\coprod_S F(\lbrace \star \rbrace),|V|)
\end{aligned}
$$

Hence $F(S) \cong \coprod_S F(\lbrace \star \rbrace)$.

With regard to the issues in the original setting about taking $\operatorname{Hom}_{\operatorname{Set}}(S,|\mathbb{Z}|)$ and then taking homs into that, there is a context in which that works: one has to work with *topological* abelian groups. Then the coproduct has the inductive topology and the product has the projective topology, and they are dual for any set regardless of size. This is, in fact, used in cohomology theories to make the duality of cohomology and homology work for non-finite CW complexes (and beyond). Or, to be more accurate, to make it even possible that cohomology is dual to homology - there are still more issues which mean that it often isn't but that is for interesting reasons rather than the dull one of failing to take topology into account. (Note that there's nothing special about abelian groups here, this works for any variety of algebras).

For more, particularly the applications to cohomology theories, see the papers by Boardman and by Boardman, Johnson, and Wilson in the *Handbook of Algebraic Topology*, and my paper with Sarah Whitehouse: *The Hunting of the Hopf Ring*.