Let $S=B\left(0,\frac{\epsilon}{2}\right)$ be a ball in $\mathbb{R}^n$. Given a translate $S+t$, there is the nice property that, for all $x,y\in S+t$, we have $x-y\in S-S\subseteq B(0,\epsilon)$. I would like to generalize this to arbitrary topological groups.
Fix a topological group $G$ with identity $e$. For each neighborhood $U$ of $e$, does there exist another neighborhood $V\subseteq U$ of $e$ such that $\{vw^{-1}:v,w\in V\}\subseteq U$?
In the above, neighborhoods are assumed a fortiori open.
I have not assumed $G$ abelian, but feel free to do so if it changes the answer.
Some observations on counterexamples
The discrete topology is not a counterexample: take $V=\{e\}$. Similarly, for the indiscrete topology, $V=U=G$.
One possible route is to look for an arithmetic obstruction. $C_3$ is not a counterexample because it is discrete, but perhaps if $\{g:\nexists x(x^2=g)\}$ is dense in $U$, then there exists no $V$? (One line of inquiry suggested by this nonexample is to considere $C_3^{\omega}$ with the product topology.)
Obviously, the argument for $\mathbb{R}^n$ goes through immediately if $G$ is locally metrizable. By Nagata-Smirnov, that means that a (hypothetical) counterexample violates regularity, Hausdorffness, or $\sigma$-local finitude in $U$.
Checking Steen & Seebach for metrization counterexamples doesn't really help: most don't have a natural group structure. The remainder are the Long Line (45), the converging rational topology (65), $\beta\mathbb{Z}$ (111), and the radial interval topology (141). All of these violate metrizability by having neighborhoods of the origin that are "too large" (i.e. not $\sigma$-locally finite), but that's the wrong place to look. If $V$ is large, then so is $U$. Instead, counterexamples should arise from groups that are not Hausdorff.
Yes, this is almost immediate from the definition of a topological group. The map $f:G\times G\to G$ defined by $f(g,h)=gh^{-1}$ is continuous (since the group product and inverse maps are continuous). Since $U$ is a neighborhood of $e$ and $f(e,e)=e$, this means $f^{-1}(U)$ is a neighborhood of $(e,e)$. That is, there exist neighborhoods $V_1$ and $V_2$ of $e$ such that $V_1\times V_2\subseteq f^{-1}(U)$. The set $V=V_1\cap V_2\cap U$ then satisfies your requirements.