Suppose $X$ and $Y$ are stably isomorphic, so that there exist a morphism $f:X\to Y$ whose image $\underline f:X\to Y$ in the stable category is an isomorphism. Then $\underline f$ has an inverse: there exists $g:Y\to X$ such that $\underline g\circ\underline f=1_X$ and $\underline f\circ\underline g=1_Y$, and this means in particular that there is a projective-injective $P$ and maps $r:X\to P$ and $s:P\to X$ such that $g\circ f-1_X=s\circ r$.

This gives us maps $F=\left(\begin{smallmatrix}f\\\a\end{smallmatrix}\right):X\to Y\oplus P$ and $G=\left(\begin{smallmatrix}g&-b\end{smallmatrix}\right):Y\oplus P\to X$ such that $G\circ F=1\_X$. *If now we assume that $\mathcal A$ has all its idempotents split*, then we can conclude that there is a $Q$ such that there is an isomorphism $H:X\oplus Q\xrightarrow{\cong} Y\oplus P$ such that $F=H\circ\iota:X\to Y\oplus P$, with $\iota:X\to X\oplus Q$ the canonical map.

Notice that $\underline F$ and $\underline H$ are isomorphisms in $\underline{\mathcal A}$, so that also $\underline\iota$ is an isomorphism there. If $p:X\oplus Q\to X$ is the projection, then $\underline p\circ\underline\iota=1\_X$ in $\underline{\mathcal A}$, so in fact $(\underline\iota)^{-1}=\underline p$, and in consequence the composition $\iota\circ p:X\oplus Q\to X\oplus Q$ is the identity of $X\oplus Q$ in $\underline{\mathcal A}$. In other words, there exists a projective-injective $R$ and morphisms $u:X\oplus Q\to R$ and $v:R\to X\oplus Q$ such that $p\circ\iota-1_{X\oplus Q}=v\circ u$.

If now $j:Q\to X\oplus Q$ and $q:X\oplus Q\to Q$ are the canonical maps, we have $q\circ v\circ u\circ j=-1\_Q$, so that the morphism $\underline{1\_Q}:Q\to Q$ is zero. This implies that in fact $Q\cong 0$ in $\underline{\mathcal A}$. By what you showed in your question, this implies that $Q$ is a projective-injective in $\mathcal A$.

All in all, we have shown that there exists projective-injectives $P$ and $Q$ such that $X\oplus Q\cong Y\oplus P$ in $\mathcal A$, as you wanted.

(I do not think your question will have a positive answer when $\mathcal A$ does not have all its idempotents split... I do not have a counterexample, though)