Let $G$ be LCA, compact, $F\subseteq G$, $F+F=G$, $F$ open. Is it true that for each $g\in G$, $g+F\cap F\neq \emptyset$? In particular, this is equivalent to $F-F=G$.
I've tried a bunch of examples for $S^1$. There seems to be a connection to the connectedness of the LCA group, but formalizing this has proved to be difficult.
It is not true when we drop the requirement that $F$ is open: consider the set $F=\{e^{2\pi i\theta}\,|\,0\le \theta<1/2\}$. Then $F+F=S^1$, but $-F\cap F=\emptyset$.
A partial result:
We can say that this is true if $G$ is connected. Let $g\in G$, and consider the cosets $\{g+F,2g+F,\ldots\}$. $\mu(F)=\mu(g+F)>0$, $\mu(G)=1$ by compactness, so for some $i\neq j$, $ig+F\cap jg+F\neq\emptyset$, and without loss of generality, $i<j$. Then $(j-i)g+F\cap F\neq\emptyset$. Let $k$ denote the minimal $k>0$ such that $kg+F\cap F\neq\emptyset$. Then the map $\theta:G\to \mathbb N$ by $g\mapsto \min\{k\,|\,kg+F\cap F\neq\emptyset\}$ is continuous: for each point $g\in G$, there is an open set $U\ni g$ such that for each $h\in U$, $\theta(h)=\theta(g)$: $\theta(g)g+F\cap F$ is a nontrivial open set and group multiplication is continuous, so the translate of a small neighborhood about the identity will suffice.
EDIT: The map $\theta$ is not continuous. Its construction didn't depend on $F+F=G$, only that $F$ was open, and any small neighborhood around the identity -- for example, $\{e^{2\pi ix}\,|\,\frac{-1}{8}<x<\frac{1}{8}\}$ -- shows that $\theta(g)$ is nonconstant, beccause $\theta(e^0)=1$, but $\theta(e^{i\pi})=2$.
Now, given that the map is continuous, note that $0+F\cap F\neq\emptyset$, and $G$ is connected, so $\theta(g)=1$ for each $g$. So $g+F\cap F\neq\emptyset$, as desired.