The map is not necessarily a weak equivalence, even if $K_i$ is actually the whole product of free groups rather than just a subgroup.
Let $K$ be the constant simplicial set which is $\Bbb Z^2$ in each degree. The map $K \to K_{ab}$ is an isomorphism, and $K = K_{ab}$ has no higher homotopy groups.
Let $G$ be a "cofibrant replacement" of $K$. This is work to describe explicitly, but there exists a version of $G$ which, in degree $n$, is free on $(n+2)$ generators. Namely, we let $G_n$ be the free group on generators $x, r_1, \dots, r_n, y$, subject to relations:
- all the face & degeneracy maps take $x$ to $x$ and $y$ to $y$,
- the generator $r_1$ in $G_1$ satisfies $d_0(r_1) = [x,y]$ and $d_1(r_1) = 1$, and
- the elements $r_i$ in $G_n$ are the images of $r_1$ under the $i$ different degeneracy maps $G_1 \to G_n$.
This forces the rest of the face and degeneracy maps by multiplicativity.
There is a map of simplicial groups $G \to K$ sending $x$ to $(1,0)$, $y$ to $(0,1)$, and $r_i$ to $0$. One can check that this is an equivalence. (It roughly corresponds to the presentation of a torus, which is a $K(\Bbb Z^2,1)$, using two 1-dimensional cells and a 2-dimensional cell.)
However, in $G_{ab}$ the first homotopy group is nonzero: the element $r_1$ becomes a cycle because its boundary $[x,y]$ became trivial, but it's not in the image of the boundary map.
All of this is an instance of an insight of Quillen's. If you take a simplicial group $K$, replace it by a weak equivalence $G \to K$ from a simplicial group $G$ that is levelwise free, and form $G_{ab}$, then the homotopy group $\pi_k G_{ab}$ is the homology group $H_{k+1}(BK;\Bbb Z)$. In this respect, group homology is a kind of derived abelianization procedure. Because, in your question, the groups $K_i$ are subgroups of products of free groups and may have nontrivial higher group homology, we can't use them as replacements for free groups when forming these resolutions.
