The answer is yes if you additionally assume $G$ is first countable.

I claim that for any open neighborhood $U$ of $0$ there exists a symmetric open neighborhood $V$ of $0$ such that $nV \subseteq U$ for all $n \in {\bf N}$. This implies that $H = \bigcup_{n \in {\bf N}} nV$ is an open subgroup of $U$, as desired. To prove the claim, suppose it fails and let $U$ be an open neighborhood of $0$ such that every symmetric open neighborhood $V$ of $0$ satisfies $nV \not\subseteq U$ for some $n$. Let $(V_i)$ be a neighborhood base at $0$ and wlog (replacing $V_i$ with $V_i \cap -V_i$) assume each $V_i$ is symmetric. Then for each $i$ we can find a finite sequence $s_i$ of elements of $V_i$ whose sum is not in $U$. Concatenating the $s_i$ yields a sequence of elements which converges to $0$, but whose sum does not converge. To see the second assertion, let $W$ be a neighborhood of $0$ such that $W - W \subseteq U$ and let $z \in G$. For each $i$ let $x_i$ be the sum of the sequence $s_1\hat{\phantom{.}}s_2\hat{\phantom{.}}\cdots\hat{\phantom{.}}s_i$. If $x_i \in z + W$ then $x_{i+1} - x_i = \sum s_{i+1} \not\in U$, so that $x_{i+1} \not\in z + W$. Thus the sequence $(x_i)$ cannot converge to $z$, and $z$ was arbitrary. This contradicts the hypothesis on $G$, so the claim is proven.