The conclusion does not follow.
Let's take
$$
G=\prod_{n=1}^\infty \mathbb{R}
$$
as a group, which we topologize by taking as a basis sets of the form
$$
\prod_{n=1}^\infty U_n,\hspace{1cm}U_n\subset\mathbb{R}, U_n\text{ open for all but finitely many $n$}.
$$
Claim: No sequence of non-zero elements of $G$ converges to 0.
Proof: Suppose $x_j=(x_{j,1},x_{j,2},\ldots)\in G$, $j=1,2,\ldots$, is a sequence of non-zero elements.
Case 1: There exists a positive integer $N$ and a finite set $S\subset\mathbb{N}$ such that $x_{j,n}=0$ whenever $n> N$ and $j\not\in S$. In this case
$$
\prod_{n=1}^N\{0\}\times\prod_{n=N+1}^\infty \mathbb{R}
$$
is an open neighborhood of 0 not containing $x_j$ when $j\not\in S$, so $x_j\not\to 0$.
Case 2: For every positive integer $N$, there exist arbitrarily large integers $j_N$ and integers $n_N>N$ such that $x_{j_N,n_N}\neq 0$. We may choose $j_1<j_2<\ldots$. Then
$$
\prod_{n=1}^\infty \bigcap_{\substack{N\\n_N=n}}\big(-|x_{j_N,n_N}|,|x_{j_N,n_N}|\big)
$$
is open (the intersections appearing are finite, and we take the empty intersection to be $\mathbb{R}$) and doesn't contain any $x_{j_N}$, so $x_j\not\to 0$. This proves the claim.
Suppose $I$ is an infinite index set and $f:I\to G$ satisfies the condition: for all open neighborhoods $U$ of $0$ in $G$, all but finitely many elements of $I$ are mapped into $U$ by $f$. The claim then implies $f(i)=0$ for all but finitely many $i\in I$ (if not, find distinct $i_1,i_2,\ldots,\in I$ not mapping to $0$, the condition implies $f(i_1),f(i_2),\ldots\to 0\in G$, contradicting the claim).
This means the sum
$$
\sum_{i\in I}f(i)
$$
is finite, hence convergent, so $G$ satisfies the hypothesis of your question.
The open neighborhood
$$
\prod_{n=1}^\infty (-1,1)
$$
of $0$ doesn't contain any open subgroups, so the conclusion is not satisfied.
