Suppose we have a countable, torsion-free abelian group $A$ with the property that for each element $a\neq 0$ the set $D\_a=\{x\in A|\exists n\in \mathbb{Z}:nx=a\}$ is finite. 
Is $A$ already a free abelian group?


If one drops the condition "countable" the infinite direct product of countably many copies of $\mathbb{Z}$ is a counterexample.



Are these abelian groups free?
No, there are non-free abelian groups of rank 2 (i.e., subgroups of $\mathbb Q^2$) in which every subgroup of rank 1 is free. (I assume you intended $a\neq 0$ in the question; otherwise the only such group would be the zero group.) In fact, such a rank-2 group can be so far from free that its quotient by any pure rank-1 subgroup is divisible. That result is due to L. Fuchs and F. Loonstra in "On the cancellation of modules in direct sums over Dedekind domains" (Indag. Math. 33 (1971) 163-169).


