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?
This is not a complete answer, but a construction that might give an answer.
I'll start by constructing a ring with several objects (a.k.a. preadditive category) $\mathcal{C}$ by generators and relations, using similar ideas as in Leavitt's ring with $R\not\cong R\oplus R\cong R\oplus R\oplus R$.
The objects $X_{1}, X_{2},\dots$ are indexed by the positive integers. For each $m\leq n$ there are generators $\alpha_{n,m},\beta_{n,m}: X_{n}\to X_{m}$ and $\gamma_{n,m},\delta_{n,m}:X_{m}\to X_{n}$.
The relations are designed to make $$X_{n}\oplus X_{n}\cong X_{1}\oplus X_{2}\oplus\cdots\oplus X_{n}$$ for all $n$ once we close under finite direct sums. For example, for $n=3$ impose the relations that make the matrices $$ \begin{pmatrix} \alpha_{3,1}&\beta_{3,1}\\\alpha_{3,2}&\beta_{3,2}\\\alpha_{3,3}&\beta_{3,3} \end{pmatrix} \text{ and } \begin{pmatrix} \gamma_{3,1}&\gamma_{3,2}&\gamma_{3,3}\\\delta_{3,1}&\delta_{3,2}&\delta_{3,3} \end{pmatrix} $$ mutually inverse.
Now form a ring by adjoining a unit: $$R=\mathbb{Z}\oplus\bigoplus_{m,n}\operatorname{Hom}_{\mathcal{C}}(X_{m},X_{n}).$$ For each $n$ let $e_{n}\in R$ be the idempotent corresponding to the identity endomorphism of $X_{n}$, let $P_{n}=e_{n}R$ be the corresponding projective right $R$-module, and let $P=\bigoplus_{n}P_{n}$.
Then $P_{n}\oplus P_{n}\cong P_{1}\oplus P_{2}\oplus\cdots\oplus P_{n}$ for each $n$, so \begin{align} P\oplus P&\cong (P_{1}\oplus P_{1})\oplus(P_{2}\oplus P_{2})\oplus(P_{3}\oplus P_{3})\oplus\cdots\\ &\cong P_{1}\oplus (P_{1}\oplus P_{2})\oplus(P_{1}\oplus P_{2}\oplus P_{3})\oplus\cdots\\ &\cong P^{(\omega)} \end{align}
Question 1. Is $P\cong P\oplus P$?
Probably not, as there seems no obvious way to produce such an isomorphism.
If the answer to Question 1 is ``no'', then this answers the supplementary question in the OP about modules.
As for abelian groups:
Clearly $R$ is countable.
Question 2. Is $R$ torsion free?
Probably, as there seems no obvious way to produce a torsion element.
Question 3. Is $R$ reduced?
Probably, as there seems no obvious way to produce a divisible element.
If the answers to Questions 2 and 3 are both ``yes'', then Corner's theorem applies, and $R\cong\operatorname{End}(B)$ for some abelian group $B$.
Let $A_{n}=e_{n}(B)$ and $A=\bigoplus_{n}A_{n}$. Then $$A\oplus A\cong A^{(\omega)}.$$
Question 4. Is $A\cong A\oplus A$?
I don't see any reason that it should have to be (note that $B$ is not uniquely determined, so the answer to Question 4 might conceivably depend on the particular choice of $B$).
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.
