FALSE: a short exact sequence splits whenever its middle object is isomorphic to the biproduct of the outer two

In $R\text{-}\mathbf{M}\mathbf{o}\mathbf{d}$, the sequence $0\to k\stackrel{i}{\to }R\stackrel{q}{\to }k\to 0$ is short exact: $q$ is the quotient by $(\epsilon )$ and $i$ identifies $k$ with that ideal. By [L1], this is a short exact sequence in an abelian category.

If this sequence split, a section $s:k\to R$ of $q$ would satisfy $q(s(1))=1$ while $R$-linearity would force $\epsilon s(1)=s(\epsilon \cdot 1)=0$, so $s(1)$ would lie in $(\epsilon )$, contradiction. Hence [L2] says the sequence is nonsplit.

Let $T:=R^{(\mathbb{N})}\oplus k^{(\mathbb{N})}$. Direct-summing step 1.1 with $0\to T\stackrel{1_T}{\to }T\to 0\to 0$ gives $0\to k\oplus T\to R\oplus T\stackrel{q\oplus 0}{\to }k\to 0$. Any section of $q\oplus 0$ would project to a section of $q$, so this stabilized sequence is still nonsplit by step 2.1.

Countable shifts give $R\oplus T\cong T$, $k\oplus T\cong T$, and $(k\oplus T)\oplus k\cong T$, because adding finitely many $R$- or $k$-summands does not change $R^{(\mathbb{N})}\oplus k^{(\mathbb{N})}$. Hence $R\oplus T\cong (k\oplus T)\oplus k$.

Step 3.1 gives a nonsplit short exact sequence, while step 4.1 shows that its middle object is abstractly isomorphic to the biproduct of its outer objects. Therefore the statement is false.