Density of “diagonal sets” in amenable groups Let $G$ be a countable amenable group with a (left) Følner sequence $(F_n)$.  Let $\Gamma$ be a subset of $G$ that has density $1$ with respect to $(F_n)$, in the sense that
$$
\lim_{n \to \infty} \frac{\lvert\Gamma \cap F_n\rvert}{\lvert F_n\rvert} \ = \ 1.
$$
Now define
$$
\Gamma' := \{ (s,t) \in G^2 : st^{-1} \in \Gamma \}.
$$
Is it true that $\Gamma'$ has density $1$ in $G^2$ with respect to $(F_n \times F_n)$?
A starting point is the calculation
$$
\lvert\Gamma' \cap F_n^2\rvert = \sum_{\gamma \in \Gamma} \#\{(s,t) \in F_n^2 : st^{-1} = \gamma \} \ = \ \sum_{\gamma \in \Gamma} \lvert F_n \cap \gamma^{-1} F_n\rvert.
$$
In the special case $G = \mathbb{Z}$, $F_n = [-n,n]$, the result is true, but my argument uses the geometry of $\mathbb{Z}$ quite strongly.  The argument is: in this case, $\lvert F_n \cap \gamma^{-1}F_n\rvert = \max(2n+1 - \lvert\gamma\rvert, 0)$, so if $n$ is chosen large enough so that $\lvert\Gamma \cap F_n\rvert > (1-\epsilon)|F_n|$, then
$$
\sum_{\gamma \in \Gamma} \lvert F_n \cap \gamma^{-1} F_n\rvert \ \geq \ \sum_{\epsilon \cdot n < \lvert\gamma\rvert \leq 2n} (2n+1-\lvert\gamma\rvert) \ \geq \ (2n+1)^2 \cdot (1 - O(\epsilon)) \ = \ \lvert F_n^2\rvert(1-O(\epsilon))
$$
as desired.
Is this result true for all countable amenable groups?
 A: The answer to your question as stated is "no", but a variant of it is true (see the proposition below).
Proof that the answer is "no": Let $(F_n)$ be the Følner sequence in $\mathbb{Z}$ given by $F_n = [2^n, 2^n + n]$, and let $\Gamma = \bigcup_{n \in \mathbb{N}}{F_n}$ so that $\Gamma$ has full density along $(F_n)$. Then $$|\Gamma' \cap (F_n \times F_n)| = \sum_{t \in F_n}{|\Gamma \cap (F_n - t)|}.$$
But for $t \in F_n$, $F_n - t \subset F_n - F_n = [-n, n]$, so
$$\frac{|\Gamma' \cap (F_n \times F_n)|}{|F_n|^2} \le \frac{|\Gamma \cap [-n,n]|}{n},$$
and it's easy to check that
$$\lim_{n \to \infty}{\frac{|\Gamma \cap [-n,n]|}{n}} = 0.$$//
The key to this counterexample is that $F_n - F_n$ behaves very differently from $F_n$ (in particular, they are disjoint sets). For the special example you considered (with $F_n = [-n,n]$), the difference set $F_n - F_n = [-2n, 2n]$ has substantial overlap with $F_n$, and that's what you were able to utilize.
For general Følner sequences, the basic idea behind your argument can still be adapted to prove the following weaker result:
Proposition: Let $G$ be a countable amenable group. If $\Gamma \subseteq G$ has full density along a Følner sequence $(F_n)$, then there exists a Følner sequence $(\Phi_n)$ in $G^2$ such that $\Gamma'$ has full density along $(\Phi_n)$.
Proof sketch: For each $n \in \mathbb{N}$, choose $N_n$ so that $F_{N_n}$ is almost-invariant with respect to shifts coming from $F_n$, say
$$\frac{|F_{N_n} \cap (F_{N_n}t^{-1})|}{|F_{N_n}|} > 1 - \frac{1}{n}$$
for $t \in F_n$.
By assumption, $|\Gamma \cap F_{N_n}| > (1 - o(1)) |F_{N_n}|$, so
$$|\Gamma' \cap (F_{N_n} \times F_n)| = \sum_{t \in F_n}{|\Gamma \cap F_{N_n}t^{-1}|} \ge |F_n| \left( 1 - o(1) - \frac{1}{n} \right) |F_{N_n}| = (1 - o(1)) |F_{N_n} \times F_n|.$$//
The Proposition can also be proved another (in my opinion, easier) way using the fact that a set has upper Banach density equal to 1 (i.e., full density along some Følner sequence) if and only if the set is thick (contains a translate of every finite set).
Proof sketch for thick sets: Assume $\Gamma$ is thick in $G$. We want to show that $\Gamma'$ is thick in $G^2$. Let $F \subseteq G^2$ be a finite set. Then let $\tilde{F} = \{st^{-1} : (s, t) \in F\}$. This is a finite set, so there exists $x \in G$ such that $x\tilde{F} \subset \Gamma$, since $\Gamma$ is a thick subset of $G$. A simple calculation shows $(x,1)F \subset \Gamma'$.
