I have come up with this diagram where the map $f : A \times B \rightarrow A$ is surjective. We define $j = coker(\pi_a, \pi_b)$. I know that $j \circ (\pi_a, \pi_b) = 0$ so that the composition of the maps on the top row is $0$ map from $A \rightarrow Z$. Commutativity gives us that traveling down through the bottom row also gives us 0 map. Since $\phi$ is the only non-identity map, $\phi$ is 0 map. We see that $j = f \circ \phi$ so that $j = 0$. Therefore, $(\pi_a, \pi_b)$ is epic.
$\require{AMScd}$ \begin{CD} A @>i_a>> A \sqcup B @>(\pi_a, \pi_b)>> A \times B @>j>> Z\\ @| @. @VfVV @|\\ A @= A @= A @>\phi>> Z \end{CD}
I am not sure that $f$ exists since Aluffi never seems to mention it in his proof in Chapter 9.1. Can we assume $f$ is the projection map $\pi_a$?