This is not really an answer, rather a longer remark about the question of the topology and the assumption that it should be Hausdorff.
Not every topology that makes a group a topological group is Hausdorff. But to any topological group we can associate a Hausdorff topological group in a canonical way. Let $G$ be a topological group and denote by $H$ the closure of $\{e\}$. Then $H$ is a normal subgroup in $G$, and the quotient group $G/H$ is Hausdorff with respect to the quotient topology. See Proposition 1-4 (vi) on page 6 of Ramakrishnan and Valenza, Fourier analysis on number field, 1999.
Based on this result, Ramakrishnan and Valenza write "Part (vi) shows that every topological group projects by a continuous homomorphism onto a topological group with Hausdorff topology. In this sense the assumption that a given group is Hausdorff is not too serious."
Nonetheless, the assumption plays an important role in Rupert's question. For, if we take the trivial topology $\mathcal{O}=\{\emptyset,G\}$, the $H=G$ and $G/H$ is the trivial group. Btw, the example of the trivial topology shows that the answer to the question (if we don't require Hausdorff) is easy: $G$ and the empty set are the only subsets that are closed in any topology on $G$ that makes $G$ a topological group.