Arithmetic progressions in finitely generated groups

Let $\Gamma$ be a free group over the alphabet $X$, $|X|\geq 2$, and put $S=X\cup X^{-1}$. Pick an increasing sequence of integers $n_i$, and put $\Lambda=\{g\in\Gamma|\exists i:\ell(g)=n_i\}$. Under some mild restrictions on the sequence $n_i$ we have that any arithmetic progression of length $\geq 3$ consists of elements of equal length. But then $a$, $ba$ and $b^2a$ have equal length. Hence $\ell(b)$ is even, and we decompose the reduced words for $a$ and $b$ as $a=vw$, $b=uv^{-1}$, where $\ell(u)=\ell(v)$, that is, $ba=uw$. In the same way $\ell(b^2a)=\ell(ba)$ implies $v=u$, thus $b=uu^{-1}=e$. But arithmetic progressions are assumed to consist of different elements, so this choice of $b$ is excluded. On the other hand the set of elements of length $=n_i$ is of positive density among all elements of length $\leq n_i$, so we obtain a set of positive upper density without arithmetic progressions of length 3.

I would assume that a similar argument works for any group of exponential word growth.

Your definition of "arithmetic progression" in a group is wrong. It should be as follows. Let $w(x,a_1,...,a_n)$ be a word. Then an arithmetic progression is the sequence $w(a_1,a_1,...,a_n), ..., w(a_n,a_1,...,a_n)$ for some a_1,...,a_n in the group (see https://en.wikipedia.org/wiki/Hales%E2%80%93Jewett_theorem). In this formulation the result is true as proved by Furstenberg and Katznelson (it was a \$100 problem from a list by Ron Graham, as far as I remember, http://www.math.ucsd.edu/~ronspubs/08_06_old_and_new.pdf).