Assuming countable choice, a real family is summable as a finite-subset net if and only if it has at most countable support and its nonzero terms are absolutely summable; its sum is independent of the enumeration

Suppose the finite-subset net converges to $L$. There are a finite $F_0\subseteq I$ and $C>0$ such that $|\sum_{i\in F}a_i|\le C$ for every finite $F\supseteq F_0$. If $P\subseteq I\setminus F_0$ is finite and all $a_i$ for $i\in P$ are positive, then $\sum_{i\in P}a_i =\sum_{i\in F_0\cup P}a_i-\sum_{i\in F_0}a_i \le C+\left|\sum_{i\in F_0}a_i\right|.$ The same argument applied to finite sets of negative terms bounds their absolute-value sums.

Conversely, let an enumeration of $S$ have absolutely convergent series sum $s$. Given $\varepsilon>0$, choose a finite initial segment $F_0$ whose remaining absolute series sum is below $\varepsilon$. For every finite $F\supseteq F_0$, $\left|\sum_{i\in F}a_i-s\right| \le \sum_{i\in S\setminus F}|a_i| <\varepsilon.$ Indices outside $S$ contribute zero, so the finite-subset net converges to $s$.

For each $n\ge1$, the sets $\{i\notin F_0:a_i^+\ge1/n\}$ and $\{i\notin F_0:a_i^-\ge1/n\}$ are finite, since a finite subset with more than $nC'$ members would have sum exceeding the bound $C'$. Every nonzero real lies in one of these level sets for some $n$ by [L4], so [L1] makes $S$ at most countable.

With any enumeration of $S$, the positive and negative partial sums are bounded by step 1.1, hence converge by [L2]. Thus the series of absolute values converges by [L3], so the enumerated nonzero terms form an absolutely convergent series.

Any two infinite enumerations differ by a bijective rearrangement, so [L2] gives the same sum; finite enumerations give the same finite-set sum by [L5]. This proves both directions and enumeration independence.