This question originates from the answer to this question: Proof model category chain complexes.
Let $i:A\to B$ be a morphism in an abelian such that $B\cong A\coprod C$ where $\coprod$ denotes the coproduct of $A$ and $C$.
Question: How does this give $B$ the structure of a product with $i$ being the inclusion of $A$ into it?
Firstly since coproducts are only defined up to isomorphism your requirement should be equivalent to saying that the triple $(B,i,j)$, where $i : A\to B$ is the morphism you began and $j : C\to B$ is composite of the coproduct inclusion $C\to A\sqcup C$ with the isomorphism $A\sqcup C\cong B$, is a coproduct. After that the universal property of coproduct produces $p: B\to A$ and $q:B\to C$ such that $pi=1_A$, $pj=0$, $qi=0$ and $qj=1_C$. The universal property of the coproduct then shows that that $ip+jq=1_B$. $B$ together with the morphisms $i,j,p,q$ satisfying the equations mentioned is what is known as a biproduct. I will leave it as an exercise to you show that for such a biproduct the triple $(B,p,q)$ is a product.