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 $B$).