What we are really working in is in the group of elements of the non-commutative formal power series ring whose constant coefficient is $1$. It contains the subgroup generated by $1+P$ and $1+Q$ which is known to be the free group on two generators. Its closure in the larger group is the pro-nilpotent completion of the free group where infinite products of commutators does make sense. In it one could have a formula of the desired kind. I think there might be results on this in for instance Lazard (who also considered the $p$-adic convergence of $((1+P)^{1/p^n}(1+Q)^{1/p^n})^{p^n}$).

To add to Pete's comments, a reference for the determination of the Schur subgroup, i.e., the set of division algebras that appear, is Yamada: The Schur subgroup of the Brauer group, SLN 397.

The case of non-perfect k is really no problem. The group ring is defined over the prime field $\mathbb Z/p$ and it modulo its radical is a product of matrix rings (by Wedderburn's theorem) with centers that are separable extensions. This implies that the same thing is true for any field of positive characteristic

Another example is that free group of finite rank $>1$ contains free subgroups of any finite (and countable) rank. Hence, one may take two free groups of different finite ranks $>1$.

Solvable groups are classical and non-solvable groups were classified by Hering (using the classification of finite simple, which is guess is OK by now). See en.wikipedia.org/wiki/2-transitive_group for reference.

Rational homotopy theory (for simply connected or just nilpotent spaces) shows exactly this dichotomy.

I thought that the proof of the strong Lefschetz for combinatorially defined intersection cohomology was proven by a reduction to the simplicial case which in turn reduces to the rational case. Hence the proof is not independent of algebraic geometry.

I don't think this works at least for $n=3$ (or $2$). We should have two elements with exactly one fixed point in common. We may assume that the common fixed point is $\infty$ and one element has $0$ as the other fixed point. Then they have the form $\mu x$ and $\lambda x+a$ (as Möbius transformations) with $a\ne0$ and $\lambda\ne 1$. Their commutator is then $x+(\mu-1)a$ which has exactly one fixed point.

I don't understand your description of Minasyan's result. A generator $x$ of the malnormal copy of $\mathbb Z$ must be conjugate to $x^{-1}$ as there are only two conjugacy classes and then it wouldn't be malnormal would it?