Seems almost correct, although admittedly it's been a bit hard to read. Careful though, the morphisms $\iota_i$ are epimorphisms (in this context, surjective) but not injective. Note that this is what you use: epis are right-cancellative.
As for the foot note, there's no need to be concerned. A slight abuse of notation is going on, if you want, think of the union of the relations as the union of the image of each relation set via the canonical inclusion of $F(A), F(A')$ to the coproduct. Techically speaking, when you identify $F(A) \ast F(A')$ with $F(A \sqcup A')$ these also carry to a different set, to which we will give the same name.
Here's the same argument, a bit more organized (if you can, draw a diagram as you read). I will note $A_1, A_2$ for the generator sets, $R_1,R_2$ the relations, $G_i = \langle A_i | R_i \rangle$ the quotients and $\pi_i : F(A_i) \to G_i$ the projections.
Let
$$
f_1 : \frac{F(A_1)}{\langle R_1 \rangle} \to H, \quad f_2 : \frac{F(A_2)}{\langle R_2 \rangle} \to H
$$
be two morphisms. They induce a pair of morphisms $f_i\pi_i : F(A_i) \to H$. Thus we have a unique morphism from the coproduct,
$$
f : F(A_1) \ast F(A_2) \to H
$$
such that $f\iota_i = f_i\pi_i$, where $\iota_i : F(A_i) \hookrightarrow F(A_1) \ast F(A_2)$ are the canonical inclusions.
Now if $r \in R_1$ is a relation, then
$$
f(r) = f(\iota_1(r)) = f_1\pi_1(r) = f_1(1) = 1.
$$
By the same argument applied to $R_2$, we see that $\langle R_1 \cup R_2 \rangle \subset \ker f$. Thus there exists a unique morphism
$$
\hat{f} : \frac{F(A_1) \ast F(A_2)}{\langle R_1 \cup R_2 \rangle} \to H
$$
such that $\hat{f}\pi = f$ with $\pi : F(A_1) \ast F(A_2) \rightarrow \frac{F(A_1) \ast F(A_2)}{\langle R_1 \cup R_2\rangle}$ the projection. We also see that $R_i \subset \ker \pi\iota_i$ and so each $\iota_i$ induces an arrow
$$
\hat{\iota_i} : \frac{F(A_i)}{\langle R_i\rangle} \to \frac{F(A_1) \ast F(A_2)}{\langle R_1 \cup R_2 \rangle}.
$$
so that $\pi\iota_i = \hat{\iota}_i\pi_i$ Now, since
$$
\hat{f}\hat{\iota_i}\pi_i = \hat{f}\pi\iota_i = f\iota_i = f_i\pi_i
$$
and each $\pi_i$ is epi, we get
$$
\hat{f}\hat{\iota}_i = f_i.
$$
Finally, the arrow $\hat{f}$ is unique in this sense: if we have another arrow $g$ such that $g\hat{\iota}_i = f_i$ for $i= 1,2$, then applying $\pi_i$ we get
$$
f_i\pi_i = g\hat{\iota}_i\pi_i = g\pi \iota_i
$$
for both $i$. However, $f$ is the unique morphism such that $f \iota_i =f_i \pi_i$ and so $g\pi = f$. But $\hat{f}$ was the only morphism so that $\hat{f} \pi = f$, hence it must be $f = g$.
This finally concludes the proof that $\frac{F(A_1) \ast F(A_2)}{\langle R_1 \cup R_2 \rangle}$ toghether with the arrows $\hat{\iota}_i : F(A_i)/\langle R_i \rangle \to \frac{F(A_1) \ast F(A_2)}{\langle R_1 \cup R_2 \rangle}$ are "a" coproduct for the objects $F(A_1)/\langle R_1\rangle$ and $F(A_1)/\langle R_2\rangle$. Since all coproducts are isomorphic, we get
$$
F(A_1)/\langle R_1 \rangle \ast F(A_1)/\langle R_2 \rangle\simeq \frac{F(A_1) \ast F(A_2)}{\langle R_1 \cup R_2 \rangle}.
$$
Now simply identify $F(A_1) \ast F(A_2) \simeq F(A_1 \sqcup A_2)$ and note that the (group generated by the) relations carry to what we have (once again) noted $\langle R_1 \cup R_2\rangle$, to obtain
$$
F(A_1)/\langle R_1 \rangle \ast F(A_1)/\langle R_2 \rangle\simeq \frac{F(A_1 \sqcup A_2)}{\langle R_1 \cup R_2 \rangle}
$$
as desired.
In a preadditive category, any finite product or coproduct is a biproduct (see here for a proof). This is not true in $\mathbf{Rng}$: the product of two rngs $A$ and $B$ is just the usual cartesian product $A\times B$, but this is typically not a coproduct under the inclusion maps $i:A\to A\times B$ and $j:B\to A\times B$ (which are the identity on one coordinate and zero on the other). In particular, these inclusions satisfy $i(a)j(b)=0$ for all $a\in A,b\in B$, so if they satisfied the universal property of the coproduct, the same would have to be true of any rng $C$ with homomorphisms from $A$ and $B$. This is clearly false in general--for instance, if $A=B=C$ is a rng with nonzero multiplication, you could take the identity homomorphisms $f:A\to C$ and $g:B\to C$ and these do not satisfy $f(a)g(b)=0$ for all $a\in A,b\in B$.
Note that more generally, since finite products or coproducts are biproducts in a preadditive category, any preadditive category with finite products or coproducts is additive, and then the addition operation on morphisms is actually uniquely determined by just the category structure (see here for instance). That is, a category with finite products or coproducts admits at most one structure of a preadditive category. So it makes sense to say such a category "is" preadditive or not, rather than saying it admits the structure of a preadditive category.
Best Answer
The inverse map $\iota$ is now not just a function, but a morphism in the category $\mathbf{Grp}$.
We already know that if $g\cdot h$ is an object in $G$, then $(g\cdot h)^{-1} = h^{-1}\cdot g^{-1}$. Because of your previous exercise (Eckmann-Hilton), this carries over to the group object: $\iota(g\circ h) = \iota(h)\circ \iota(g)$.
Now because $\iota$ is a group morphism, we also have $\iota(g\cdot h) = \iota(g)\cdot \iota(h)$. This additional property establishes that $\iota(g)\circ \iota(h) = \iota(gh) = \iota(h)\circ \iota(g)$, which proves commutativity.
Your proof was entirely correct; you just didn't use the fact that $\iota$ is a group morphism, which means that (because the multiplication structure is the same) the group object, and hence the original group, must be commutative. (Noncommutative groups don't have corresponding group objects because their inversion operator fails to be a homomorphism of groups.)