I should make my comments an answer.
- An open subgroup $H$ of a topological group is closed: Its complement is the open set $\bigcup\limits_{g \notin H} gH$.
- For every neighborhood $U$ of the identity element of a topological group the subgroup $H = \bigcup_{n \in \mathbb{Z}} U^{n}$ is open (hence closed by 1.): for every $h \in H$ the set $hU$ is contained in $H$ and a neighborhood of $h$.
- If $G$ is connected and $U$ is a neighborhood of the identity the subgroup $H$ in 2 is open and closed, hence the entire group $G$.
This implies that a connected locally compact group $G$ is $\sigma$-compact (since we may take a compact neighborhood $U$ of the identity in the above), hence every Radon measure is $\sigma$-finite and thus there are no examples as you're asking for: if $S \subset G$ is any measurable subset then $\{0\} \times S$ is a measurable rectangle, hence measurable in $\mathbf{R} \times G$ and by Fubini-Tonelli it must have measure zero.
If connectedness is dropped then this Fubini-argument fails. The standard example is $\{0\} \times \mathbf{R}_d$ in $\mathbf{R} \times \mathbf{R}_d$ where $\mathbf{R}_d$ is the additive group of the reals equipped with the discrete topology and $\mathbf{R}$ carries the standard topology. The set $\{0\} \times \mathbf{R}_d$ is locally null, but not null (hence it has infinite measure). This is an exercise all mathematicians interested in analysis should do once in their lives, so I won't spell it out. In case of emergency consult (11.33) on p.127 in Hewitt-Ross, Abstract Harmonic Analysis, I.