Update: Steve points out in comments that many direct products of residually cyclic groups will be counterexamples. Indeed, I should have spotted this: every finitely generated abelian group is residually cyclic! I should have asked about fully residually cyclic groups, ie groups $G$ such that for any finite subset $X\subseteq G\smallsetminus 1$, there is a homomorphism from $G$ to a cyclic group that doesn't kill any elements of $X$. I suspect the question is now quite easy, though I haven't had a chance to think about it yet.
This question is motivated by this recent question, which asked for a characterisation of groups with at most one subgroup of each finite index. Arturo Magidin's answer showed that every such finite group is cyclic, but the questioner, Louis Burkill, added in comments that he is really interested in the infinite case.
In my answer to that question, I argue that a finitely generated group has at most one subgroup of each finite index if and only if its canonical residually finite quotient, $R(G)$, is cyclic. The `finitely generated' hypothesis is necessary - otherwise the additive group of the rationals provides a counterexample.
In the general case, one can reduce from the case of $G$ to $R(G)$ as before (this gets rid of pathological examples like infinite simple groups), and the same argument shows that if $R(G)$ has at most one subgroup of each finite index then $R(G)$ is fully residually cyclic, in particular abelian. But the converse is not clear to me. Hence the question in the title, which I'll reiterate here, with some extra hypotheses that should indicate where the difficulty lies.
If $G$ is an infinitely generated, fully residually cyclic (in particular, abelian) group, must $G$ have at most one subgroup of each finite index?