The union of the two groups involved in a free product is not quite disjoint $-$ they have to share the same identity element. But otherwise yes, $G*H$ is the free group on $G\cup H$ modulo $1_G=1_H$ and the multiplication tables already in place for $G$ and $H$. Indeed if $G=\langle X|R\rangle$ and $H=\langle Y|S\rangle$ are presentations with $X,Y$ disjoint (by fiat) then $G*H=\langle X\cup Y|R\cup S\rangle$.
In the category of abelian groups the coproduct is the direct sum. This can be obtained from the free product by imposing commutativity: (assuming $G,H$ commutative) we can interpret $G,H$ themselves as subgroups of $G*H$, and we have $G\oplus H= (G*H)/[G,H]$. This is indeed the most free abelian group generated by $G\cup H$ (modulo $1_G=1_H$).
As k.stm says in the comments, usually a more general thing is true: these categories $C$ are equipped with forgetful / underlying set functors $U : C \to \text{Set}$ which tend to have a left adjoint, the "free" functor $F : \text{Set} \to C$. Whenever this is true, it follows that $U$ preserves all limits, not just products.
Sometimes, but more rarely, $U$ will also have a right adjoint, which will give a "cofree" functor. Whenever this is true, it follows that $U$ preserves all colimits, not just coproducts. This happens, for example, when $C = \text{Top}$: here the left adjoint equips a set with the discrete topology and the right adjoint equips a set with the indiscrete topology. But it doesn't happen for, say, groups or rings.
So one way to reformulate your question is:
Why do the forgetful functors $U : C \to \text{Set}$ we write down tend to have left adjoints, but not right adjoints? Equivalently, why are there usually free structures, but not usually cofree structures?
A rough answer is that we can expect "free" structures whenever the structure is described by operations satisfying equational axioms, since then we can build free objects by applying all possible operations modulo all axioms. On the other hand, operations also make it difficult for the forgetful functor to preserve coproducts, since in a coproduct of two structures you can apply operations to elements of both structures, so you'll usually get something bigger than the disjoint union.
(Dually, you should expect "cofree" structures whenever the structure is described by "co-operations," and this does in fact happen: for example, the forgetful functor from coalgebras to vector spaces has a right adjoint but not a left adjoint, called the cofree coalgebra.)
A more precise answer would invoke, say, the machinery of Lawvere theories, which among other things has the benefit of also telling you exactly what colimits you can expect these forgetful functors $U$ to preserve. This is a long story so I don't want to get into it unless you feel like it really answers your question, but the gist is that Lawvere theories present familiar structures like groups, rings, and modules in a way that fundamentally uses finite products, but nothing else. You can deduce from this that the forgetful functor $U$ preserves (and in fact creates) any limits or colimits that commute with finite products in $\text{Set}$. Every limit commutes with finite products, and the colimits that commute with finite products in $\text{Set}$ are precisely the sifted colimits. These include, for example, increasing unions, which is an abstract way to see why the set-theoretic increasing union of a sequence of groups is still a group, and the same with groups replaced by rings, modules, etc.
Best Answer
Coproducts in the category of preadditive categories are just disjoint unions with a zero morphism added between each pair of objects from the different original categories. Indeed, if $C$ and $D$ are preadditive categories, then the category $E$ obtained in this way is preadditive, and a pair of additive functors out of $C$ and $D$ extends uniquely to an additive functor out of $E$ by just sending each of the new zero morphisms to zero. It follows that $E$ satisfies the universal property of the coproduct in the category of preadditive categories.
Coproducts do not exist in the category of additive categories. For instance, there is not even an initial object. The issue is that if $C$ is a category in which every object is a zero object but $C$ has many different objects, then every additive category $D$ has many different functors to $C$ (one for every function $\operatorname{Ob}(D)\to \operatorname{Ob}(C)$).
However, coproducts do exist in the 2-category of additive categories, and are the same as products. Coproducts in a 2-category are defined to only have their universal property "up to isomorphism". To be more precise, a coproduct in a 2-category of objects $C$ and $D$ is an object $E$ with morphisms $i:C\to E$ and $j:D\to E$ such that composition with $i$ and $j$ gives an equivalence of categories $\operatorname{Hom}(E,F)\simeq \operatorname{Hom}(C,F)\times \operatorname{Hom}(D,F)$ for any object $F$.
The idea behind coproducts in the 2-category of additive categories being the same as products is that an object $(c,d)$ of the product represents the formal direct sum $c\oplus d$. Explicitly, let $C$ and $D$ be additive categories. Then there is an additive inclusion functor $i:C\to C\times D$ taking $c\in C$ to $(c,0)$ for some chosen zero object $0\in D$, and similarly there is an additive inclusion functor $j:D\to C\times D$. Now given any other additive (or abelian) category $E$ and additive functors $f:C\to E$ and $g:D\to E$, we can define an additive functor $h:C\times D\to E$ by $h(c,d)=f(c)\oplus g(d)$. This satisfies $hi=f$ and $hj=g$, and it is easy to see that it is the unique such $h$ up to natural isomorphism (essentially because $(c,d)=(c,0)\oplus(0,d)$ in $C\times D$). With a bit of work we can show that this in fact gives an equivalence between the category of additive functors $C\times D\to E$ and the category of pairs of additive functors $C\to D,C\to E$ and thus $C\times D$ is a coproduct of $C$ and $D$ in the 2-category of additive categories.
The story for abelian categories is exactly the same as for additive categories.
As a final aside, you will notice that these coproducts have nothing to do with tensor products. That's because tensor products are about multiplication, but when you combine two (pre-)additive categories, there's no way in which you need to be able to "multiply" morphisms from them. Instead, you just need to be able to take direct sums of objects (to get an additive category). This is similar to how coproducts of abelian groups are direct sums, not tensor products, since there is no multiplication operation for abelian groups. (On the other hand, coproducts of commutative rings are tensor products, since they require a multiplication operation).