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).
I feel like the key observation here is that in some categories (like groups) we have a canonical map from coproducts to products, induced by these inclusion maps that you've noticed exist. The natural question is where does it come from? Perhaps answering this question will shed some light on what's going on here, since we can then take a look at other examples of this phenomenon to get intuition.
Motivation: (This section is sorta fuzzy mathematically in places to motivate the definitions without worrying about details)
So what is a map from the coproduct to the product? Well by definition, if $\newcommand\C{\mathcal{C}}\C$ is our category, then
$$
\newcommand\of[1]{\left({#1}\right)}
\C\of{
\coprod_i X_i,
\prod_j X_j
}
\cong
\prod_{i,j}\C(X_i,X_j).
$$
So a map from the coproduct to the product requires for every pair of objects
a choice of map $\C(X,Y)$. When $X=Y$, this is easy, we can take the identity map.
What do we do if $X\ne Y$ though? Well, if we require our category has finite products/coproducts including initial and terminal objects, $0$ and $1$ respectively, then if the unique map $0\to 1$ is an isomorphism, then we can always produce a map
$$X\to 1\to 0\to Y,$$
and this doesn't depend on our choice of $0$ or $1$, since everything is unique up to isomorphism.
This gives us a definition.
Categories with zero objects:
If $0\to 1$ is an isomorphism, then $0$ is both initial and terminal, and we say $\C$ has a zero object, and from now on I'll write $0$ for a zero object.
We also say that the unique map $X\to 0\to Y$ is the zero morphism from $X$ to $Y$, written as $0$. (This gives a canonical enriching in pointed sets with smash product, which is related to your observation that groups have a distinguished element).
Thus in a category with zero objects, we can define a canonical morphism
$$
\coprod_i X_i \to \prod_j X_j,
$$
with components $1_{X_i}$ when $i=j$ and $0$ when $i\ne j$. (Writing this as a matrix, you'll note that this is the identity matrix).
However, this canonical map is not generally an isomorphism. When it is (for finite sums/products), we call the object the biproduct, written $X\oplus Y$, and in such a situation, we get a canonical enriching in commutative monoids. The addition of $f,g : X\to Y$ is given by the composite
$$ X\xrightarrow{\Delta} X\oplus X \newcommand\toby\xrightarrow\toby{f\oplus g}
Y\oplus Y \toby{\nabla} Y. $$
Examples:
The trivial group, $1$, is the zero object in both groups and abelian groups, and you can check that the canonical morphism I define above gives you the same result as the morphism induced by the universal property given by your inclusions. I.e., it sends
$g \in G_i$ to the tuple $(1,1,\ldots,1,g,1,\ldots,1)$, with $g$ in the $i$th spot.
For another category, you have the category of pointed sets or pointed topological spaces (pairs $(X,x)$ with $x\in X$ and morphisms $f:(X,x)\to (Y,y)$ are the maps $f:X\to Y$ such that $f(x)=y$).
The coproduct here is called the wedge sum, and it naturally includes into the product in the same way, with $x\in X_i$ mapping to
$(*,\ldots,*,x,*,\ldots,*)$, where $x$ is in the $i$th location, and
$*$ is the basepoint in the other factors.
Finally, let's give a bit of a weird one. (Though this is a category I came across recently.)
$R$-algebras (for me right now, the category of commutative, unital algebras over a commutative ring $R$) do not have zero objects (unless $R=0$). The initial object is $R$, and the terminal object is the zero ring. However, we can consider the category of
$R$-algebras with augmentations. Explicitly, these are commutative rings $S$ with
maps
$$ R\toby{\iota_S} S \toby{\pi_S} R,$$
where $\pi_S\iota_S = \newcommand\id{\mathrm{id}}\id_R$.
Morphisms are ring maps $\phi : S\to T$ such that $\pi_T\phi = \pi_S$ and
$\phi\iota_S = \iota_T$. Now $R$ is a zero object in this category. (This is a special case of a general way to produce new categories with zero objects, make the objects either pairs $(X,1\to X)$ of an object and a morphism from the terminal object or the dual construction, take pairs $(X,X\to 0)$, which is what we did here). The first construction is common, and called taking the pointed category of $\C$, denoted $\C_*$, and is what we do to produce pointed sets and pointed topological spaces.
The coproduct of $S$ and $T$ is $S\otimes_R T$, with augmentation given by $(s\otimes t)\mapsto \pi_S(s)\pi_T(t)$. The product of $S$ and $T$ is given by (the fiber product) $S\times_R T$, with algebra structure map given by $r\mapsto (\iota_S(r),\iota_T(r))$.
Then the morphism $S\otimes_R T\to S\times_R T$ is given by $s\mapsto (s,\iota_T(\pi_S(s))$ and $t\mapsto (\iota_S(\pi_T(t)),t)$.
Conclusion
Hopefully I've given a bit of context and background to the construction that you've noticed. I hope you'll see that asking the question of why the product fails to be the coproduct in groups, while a reasonable question, doesn't have much of an answer other than: because it can't be.
While that can feel very unsatisfying as an answer, I hope that seeing that there are lots of similar examples will make the following analogy make sense.
The inclusions produce a map from the coproduct to the product, in my analogy, I want to think that this is like proving an inequality. And sometimes this inequality is a strict inequality (the map is not an isomorphism), but sometimes in special cases, the inequality is an equality (the map is an isomorphism), and then something special and interesting happens. But the point is that asking why something is a strict inequality is hard to answer with anything other than because we can prove that they aren't the same. Instead, asking when do we have equality may be more fruitful. (Not a perfect analogy, granted.)
Best Answer
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 $1_G=1_H$).