Since you mentioned in the comments you have, and are familiar with Topology by Munkres, and I think Munkres takes a similar approach to what your question is discussing, I will use it as a guide, and explain some of what is going on there.
So first off, to actually construct an example of a free product of groups, typically you go through this reduced word definition, in which case you need disjoint groups, otherwise there is interaction between subgroups. This is purely convenient for construction, and not necessary for definition of free products (although this is actually what Munkres takes as the definition of free product of groups).
Munkres then defines what it means to be an external free product of $G_\alpha$ relative to monomorphisms $i_\alpha:G_\alpha \to G$ to be when $G$ is a free product of $i_\alpha(G_\alpha)$. Note that he does not require $G_\alpha$ to be disjoint, but by his definition of free product the $i_\alpha(G_\alpha)$ must be disjoint. To show that given $\{G_\alpha\}$ that an external free product always exist, what Munkres does is move to assuming that they are disjoint, which can be done by passing to $\{G_\alpha \times \{\alpha\} \}$ and doing the word construction. Taking this route there are natural monomorphism $j_\alpha:G_\alpha \times \{\alpha \} \to G$, so $G$ is an external free product of this collection, and there are natural isomporphism $\iota_\alpha:G_\alpha \to G_\alpha \times \{\alpha \}$, so $i_\alpha= j_\alpha \iota_\alpha$ is a collection of monomorphisms, and $G$ is in fact an external free product of the $G_\alpha$, and you don't really have to assume the $G_\alpha$ are disjoint, it is just helpful for the actual construction. This "little trick" is frequently used in other places (basically whenever you are constructing things from maps satisfying certain properties), and I think it is one of those things that Munkres, and others would probably expect you to fill or "just know" how to do/it doesn't effect anything.
In a couple of the other answers here, free products is defined in terms of universal property (Munkres uses the term extension condition, or something similar), and note that this does not assume things are disjoint either, for essentially the same reasons as why you don't require the external free product to not be disjoint.
In a little bit more concrete example, if we want to think about $\mathbb{Z} * \mathbb{Z}$ (which is just a name for a group satisfying some universal property) in terms of external free product, we get that there are monomorphisms $i_1,i_2: \mathbb{Z} \to \mathbb{Z} * \mathbb{Z}$ such that $i_1(\mathbb{Z}) \cap i_2(\mathbb{Z})= \{id\}$, and it is a free product $i_1(\mathbb{Z})*i_2(\mathbb{Z})$ in the reduced word sense.
If you look in Munkres I don't think he ever writes anything like $\mathbb{Z} * \mathbb{Z}$, basically because it disagrees with his approach, but from the above example it is pretty easy to see what that would mean, no matter what approach you take to understanding free products.