In the comments to this answer, Martin Brandenburg told me that it is possible to derive the element structure of the coproduct $A\oplus B$ of two abelian groups $A$ and $B$, by knowing that such a coproduct exists and by using the universal property. In particular, it is not necessary to use an explicit description of it. Since I usually have to deal with situations in which I know the existence of a universal solution to a certain question, but I don't know how to explicitly describe it, I would be glad to learn how this could be done in this case.
For me, given two abelian groups $A,B$ their coproduct is an abelian group $Z$ together with two group homomorphisms $j_A:A\to Z$ and $j_B :B\to Z$ which is universal with respect to this property. That is to say, such that for any other abelian group $C$ together with two morphisms of abelian groups $f:A\to C$ and $g:B\to C$, there exists a unique morphism of abelian groups $\nabla(f,g):Z\to C$ such that $\nabla(f,g)\circ j_A = f$ and $\nabla(f,g)\circ j_B = g$.
I tried to solve the exercise Martin left to me. First of all, since we are in the category of abelian groups, we have the zero morphism and hence I can consider $\nabla(id_A,0):Z\to A$ and $\nabla(0,id_B):Z\to B$.
Now I need to assume that I know that products in the category of abelian groups exist and that they can be realized as the product of the underlying sets together with the component-wise composition law. Thanks to these assumptions, I know that there exists a unique group homomorphism $\Phi:Z \to A\times B$ such that $\pi_A\circ\Phi=\nabla(id_A,0)$ and $\pi_B \circ \Phi = \nabla(0,id_B)$. By composing, I can consider $i_A:=\Phi\circ j_A:A\to A\times B$ and $i_B:=\Phi\circ j_B :B \to A\times B$. Since $\pi_A\circ\Phi\circ j_A = id_A$ and $\pi_B\circ\Phi\circ j_B = id_B$, both $i_A$ and $i_B$ are injective (and $j_A$ and $j_B$ as well).
At this point I stopped because I realized that I am not "describing the element structure" of $A\oplus B$. I am showing that $A\oplus B$ is isomorphic to $A\times B$ with component-wise composition law (next step would have been to use the universal property of $A\times B$ to show that there exists a unique group homomorphism $\Psi:A\times B\to Z$ such that $\Psi \circ \Delta(id_A,0) = j_A$ and $\Psi \circ \Delta(0,id_B) = j_B$ and then proving that, by uniqueness and the explicit description of $A\times B$, $\Phi$ and $\Psi$ are each other inverses). Thus, I am providing an explicit realization of $A\oplus B$, which is not what has been asked me.
Can anybody suggest me a different approach?