$A^2$ is isomorphic to $A^{(\omega)}$, but not $A$ Is there an abelian group $A$ with $A\not\cong A\oplus A\cong A\oplus A\oplus A\oplus\cdots$ (a direct sum of countably many copies of $A$)?

Edited to add: As no answers are forthcoming, does anyone know what happens if we allow arbitrary modules in place of abelian groups?
 A: This is a long comment responding to Simon Henry's question in the comments to the original question.
The paper
A.L.S. Corner
On a conjecture of Pierce concerning direct decompositions of Abelian groups
Proc. Colloq. Abelian Groups, Tihany, 1963, Akadémiai Kiadó, Budapest (1964), pp. 43–48
provides an example of an abelian group $G$ such that $G\cong G\oplus G\oplus G$ and $G\not\cong G\oplus G$. I don't have access to the paper, so can't say whether Corner's example will solve the original question on this page. Here are the first two sentences of the Math Review, written by R. S. Pierce:
 It is shown that for any positive integer $r$ there exists a countable torsion-free abelian group $G$ such that the direct sum of $m$ copies of $G$ is isomorphic to the direct sum of n copies of $G$ if and only if $m\equiv n\pmod{r}$. This remarkable result is obtained from the author's theorem on the existence of torsion-free groups having a prescribed countable, reduced, torsion-free endomorphism ring by constructing a ring with suitable properties.
