$\def\colim{\operatorname{colim}} \def\Ker{\operatorname{Ker}}$This is not a stupid question. First of all, he says AB3 should be satisfied and the thing you say. In particular Grothendieck's AB5 implies the existence of colimits. Now let's for example try to show that colimits over $\mathbf{N}$ are exact if Grothendieck's AB5 holds. So assume given a short exact sequence of systems $0 \to (A_n) \to (B_n) \to (C_n) \to 0$ over $\mathbf{N}$. It is clear that $\colim A_n \to \colim B_n \to \colim C_n \to 0$ is exact by looking at Homs into another object and using the mapping property of $\colim$. OK, now let $K \subset \colim A_n$ be the kernel of the first map. Let $A'_n \subset \colim A_n$ be the image of $A_n$. Observe that $\colim A_n = \sum A'_n$ as the map $\bigoplus A_n \to \colim A_n$ is surjective. Then Grothendieck's axiom AB5 says it suffices to show that $K \cap A'_n = 0$. I claim that $\Ker(A_n \to A'_n) = \bigcup \Ker(A_n \to A_{n + m})$. To see this I suggest you think about the exact sequence
$$0 \to \bigoplus A_n \to \bigoplus A_n \to \colim A_n \to 0$$
and use Grothendieck's axiom AB5 about unions inside the object in the middle. A similar argument shows that the inverse image of $K$ in $A_n$ is $\bigcup \Ker(A_n \to B_{n + m})$. Since the maps $A_{n + m} \to B_{n + m}$ are injective we conclude. Presumably there are shorter proofs as well.

Anyway, it doesn't matter as we will use AB5 as stated throughout the document.