Let $G$ be an abelian group with a presentation of abelian group given only by relators of the form $g^n=h$, $g,h$ generators and $n$ integer. Then $G$ is a direct sum of cyclic groups. We can assume we allow relators $g^n=1$ (because it can be encoded in adding a generator $k$ and adding the relators $h=k$ and $h^2=k$). Now given such a presentation (say "satisfying ($*$)") we can remove all generators representing the identity (replacing them with 1 in relators) and the resulting presentation still satisfies ($*$) and has the additional property that no generator maps to the identity. Now consider the equivalence relation on the set of generators  generated by $g\simeq h$ if the (nontrivial) generators $g,h$ occur in a single relator. If $C$ is an equivalence class, consider the group $G_C$ with the presentation with generators in $C$ and relators involving generators of $C$. Then $G$ is the coproduct of the $G_C$. Note that so far this does not make use of the torsion-freeness of $G$, and the argument works for modules over an arbitrary ring.

Now we are reduced to understand the case when the equivalence relation defined above has a single equivalence class (i.e., is the undiscrete one). Then assuming $G$ torsion-free results in the fact that $G$ is locally cyclic (i.e., for a torsion-free group, isomorphic to a subgroup of $\mathbf{Q}$).

Thus in general, assuming that the abelian group $G$ is torsion-free implies that $G$ is a direct sum of locally cyclic groups (and more generally, in the category of $A$-modules when $A$ is a domain, assuming that $M$ is torsion-free implies that $M$ is a direct sum of locally cyclic modules).

Among torsion-free abelian groups, those that are direct sums of locally cyclic groups are pretty rare. For instance $\mathbf{Z}_p$ does not satisfy this property, as any torsion-free abelian group $A$ of $\mathbf{Q}$-rank $\ge 2$, not containing any copy of $\mathbf{Z}[1/p]$ and such that $A/pA$ is cyclic); many subgroups of $\mathbf{Z}[1/p]^2$ are ruled out by this criterion. Also any non-free subgroup of $ \mathbf{Z}^X$ fails to be such a direct sum; this includes $\mathbf{Z}^{X}$ itself when $X$ is infinite.