Kernel of $G\ast H\to G\times H$ is free Let $G$ and $H$ be groups. The identity homomorphism $G\to G$ and the trivial homomorphism $H\to G$ give a homomorphism $G\ast H\to G$ by the universal property of the coproduct. Similarly, we obtain a homomorphism $G\ast H\to G$. Then the universal property of the product gives a homomorphism $G\ast H\to G\times H$. We can consider the kernel $K$ of this homomorphism. It turns out that $K$ is a free group ($K$ is freely generated by commutators $[g,h]$ for $g\in G\setminus\{1\}$ and $h\in H\setminus\{1\}$).
The construction of $K$ was very category-theoretic and only required the use of the universal properties of the product and coproduct. Similarly, the conclusion that $K$ is free is also a category-theoretic statement. Is there a proof that $K$ is free that is "category-theoretic" and avoids directly showing that the commutators $[g,h]$ for $g\in G\setminus\{1\}$ and $h\in H\setminus\{1\}$ freely generate $K$?
As people are mentioning in the comments, it is unlikely for such a proof to exist do to the reliance on a choice of generators in the categorical definition of a free group. Is there a (nonabelian) category similar to the category of groups in which this property fails?
 A: The result is not "categorical". There are categories, even varieties of groups, where it fails.
Consider the variety of groups satisfying the laws $x^4=1$, $[x,y]^2=1$, and $[x,y,z]=1$. This is the variety of nilpotent groups of class $2$, exponent $4$, and all commutators of exponent $2$.
Let $G=\langle x\mid x^2=1\rangle$, $H=\langle y\mid y^2=1\rangle$. Then their free product in this variety is their $2$-nilpotent product, which consists of all elements of the form $x^ay^bz^c$, where $0\leq a,b,c<2$, and multiplication is given by $(x^ay^bz^c)(x^{\alpha}y^{\beta}z^{\gamma}) = x^{a+\alpha} y^{b+\beta} z^{c+\gamma+b\alpha}$, with exponents taken modulo $2$.
The kernel $K$ of the map $G\amalg H\to G\times H$ is generated by $z$, and so is cyclic of order $2$. However, the free group of rank $1$ in this variety is cyclic of order $4$, not $2$, so this kernel is not free.

For an example where what you get is not a subgroup of the corresponding free group, take the variety of all nilpotent groups of class $2$. If you take $G$ and $H$ each to be the relatively free group of rank $2$ (this is isomorphic to the group of $3\times 3$ upper triangular matrices with coefficients in $\mathbb{Z}$ and $1$s in the diagonal), then their free product is again their $2$-nil product (similar to the above, the elements are of the form $ghz$ with $g\in G$, $h\in H$, and $z\in[H,G]\cong H^{\rm ab}\otimes G^{\rm ab}$, with product $(ghz)(g’h’z’) = (gg’)(hh’)(zz’[h,g’])$). Here, the kernel of the map is isomorphic to $\mathbb{Z}^4$, but the free groups of rank greater than $1$ are not abelian.
