What subsets of $\Bbb R$ are closed under countable sums?

Not quite what you want, but too long for a comment:

Proposition. If $S$ is a subgroup, then $S$ is good iff it is closed.

Proof. Suppose that $S$ is good and consider a converging sequence $s_n \to s$ with each $s_n \in S$; setting $t_n := s_{n+1}-s_n \in S$ we obtain

$$S \ni \sum_{n \geq 1}t_n = \lim_{n\to \infty}s_{n+1} = s-s_1.$$

Thus $s = (s-s_1) + s_1 \in S$. Conversely, suppose that $S$ is good and take a converging series $\sum_{n \geq 1}s_n$ with terms belonging to $S$. Since each partial sum is in $S$, the series is a limit point of $S$ (and thus belongs to $S$). $\square$

A closed subgroup of $\Bbb R$ is of the form $\alpha \Bbb Z$ for some $\alpha > 0$, $\Bbb R$ or $\{0\}$, see here.

Also, a remark: by Riemann's rearranging theorem any good subset that contains a sequence whose series is conditionally convergent must be $\Bbb R$.

Maybe if we consider the subgroup generated by a good subset $S$ we can get some more information. It is not obvious to me whether this remains to be a good subset, though.

Write $S^{\times} = S\setminus\{0\}$. Then we will prove the following claim:

Claim. A subset $S$ of $\mathbb{R}$ is good if and only if $S$ satisfies one of the following options:

1. $S$ is one of $\varnothing$, $\{0\}$, $(0, \infty)$, $[0, \infty)$, $(-\infty, 0)$, $(-\infty, 0]$, or $\mathbb{R}$.
2. $S^{\times}$ is a semigroup contained in $[r, \infty)$ for some $r > 0$.
3. $S^{\times}$ is a semigroup contained in $(-\infty, -r]$ for some $r > 0$.
4. $S = \alpha \mathbb{Z}$ for some $\alpha > 0$.

Proof. It is obvious that a set $S$ satisfying one of the above options is always good. For the opposite direction, let $S$ be a good subset of $\mathbb{R}$. Then both $S^+ = S \cap (0, \infty)$ and $S^- = S \cap (-\infty, 0)$ are good, since they are intersections of good sets. Then we have the following trichotomy for $S^+$:

\begin{equation*} S^+ = \varnothing, \quad\text{or}\quad S^+ = (0, \infty), \quad\text{or}\quad \inf S^+ > 0, \end{equation*}

and similarly for $S^-$:

\begin{equation*} S^- = \varnothing, \quad\text{or}\quad S^- = (-\infty, 0), \quad\text{or}\quad \sup S^- < 0. \end{equation*}

Now let us examine all the nine possibilities:

$$ \begin{array}{c|ccc} & S^+ = \varnothing & S^+ = (0, \infty) & \inf S^+ > 0 \\ \hline S^- = \varnothing & S^{\times} = \varnothing & S^{\times} = (0, \infty) & \text{Option 2} \\ S^- = (-\infty, 0) & S^{\times} = (-\infty, 0) & S = \mathbb{R} & \text{impossible} \\ \sup S^- < 0 & \text{Option 3} & \text{impossible} & \text{Option 4} \\ \end{array} $$

Here we only cover non-trivial cases:

• Suppose $S^+ = (0, \infty)$ and $\sup S^- < 0$. Then by picking $a > 0$ so that $-a \in S^-$, we find that $(-a, \infty) = (-a) + S^{+} \subseteq S$. This contradicts $\sup S^- < 0$.

• Suppose $\inf S^+ > 0$ and $\sup S^- < 0$. In this case, we claim:

  Claim. Whenever $a \in S^+$ and $b \in S^-$, we have $a/b \in \mathbb{Q}$.

  Let us first check that this claim implies the assertion in Option 4. For any $a \in S^+$ and $b \in S^-$, write $a/b = -p/q$ for some $p, q \in \mathbb{Z}^{>0}$ in lowest term and write $g = a/p = -b/q$. Then by the Bezout's identity, we can find integers $x, y$ such that $xp - yq = 1$. By replacing $x$ and $y$ by $x+qk$ and $y+pk$ if necessary, we may assume that both $x$ and $y$ are positive. Then $xa + yb = g \in S$. A similar reasoning also shows that $-g \in S$. So the good set generated by $\{a, b\}$ is precisely $g\mathbb{Z}$.

  Next, it is easy to see that the claim implies that both $S^+$ and $S^-$ are countable. So we may enumerate them by $S^+ = \{a_1, a_2, \dots\}$ and $S^- = \{ b_1, b_2, \dots\}$. Then the above observation yields $$\langle \{a_1, b_1, a_2, b_2, \dots, a_n, b_n\}\rangle = g_n \mathbb{Z}$$ for some $g_n > 0$ such that $g_n/g_{n+1} \in \mathbb{Z}$ for all $n$. Moreover, the assumption tells that $(g_n)_{n\geq 1}$ eventually stabilizes with some value $\alpha > 0$. Therefore $S = \alpha \mathbb{Z}$.

  So it remains to prove the claim. Suppose otherwise. Then there exist $a \in S^+$ and $b \in S^-$ such that $a/b \notin \mathbb{Q}$. Write $\xi = -a/b$ and note that $\xi$ is irrational. So by the Dirichlet approximation theorem, for any $\epsilon > 0$ we can find positive integers $x, y$ such that $0 < \left| x \xi - y \right| < \epsilon/|b|$. This then implies that $0 < \left| xa + yb \right| < \epsilon$, hence contradicts the assumption that $\inf S^+ > 0$ and $\sup S^- < 0$. Therefore the claim must be true.