Firstly I shall prove that indeed the rearrangements $R$ that preserve the convergence and value of every series form a group.$\def\nn{\mathbb{N}}$ Every rearrangement can of course be considered as a permutation on $\nn$. Every sequence is also just a function on $\nn$. We shall use $\sum^\infty x$ to denote the limit of the partial sums of $x$ if it exists and $null$ otherwise.

For any $f \in R$:

$\sum^\infty a = \sum^\infty ( a \circ f )$ for any sequence $a$:

For any sequence $a$:

$\sum^\infty ( a \circ f^{-1} ) = \sum^\infty ( a \circ f^{-1} \circ f ) = \sum^\infty a$.

Therefore $f^{-1} \in R$.

For any $f,g \in R$:

$\sum^\infty a = \sum^\infty ( a \circ f )$ for any sequence $a$:

$\sum^\infty a = \sum^\infty ( a \circ g )$ for any sequence $a$:

For any sequence $a$:

$\sum^\infty a = \sum^\infty ( a \circ f ) = \sum^\infty ( a \circ f \circ g )$.

Therefore $f \circ g \in R$.

[Okay I did look at the paper (**and I hate pay-walls for mathematics papers**) and the collection that they say is not a group is the permutations that make a convergent series into another convergent one. The paper even mentions that the collection you are interested in is obviously a group.]

A sufficient condition for the rearrangement sequence to be sum-preserving is that there is a natural $k$ such that for any natural $n$ no more than $k$ numbers in $[1..n]$ are missing from the length-$n$ prefix of the sequence.

Clearly if the above condition holds, then the discrepancy between the partial sums of the original and rearranged series are bounded by the sum of magnitude of at most $2k$ terms. These terms have eventually increasing indices, because the numbers in $[1..n]$ that are not in the length-$n$ prefix will eventually appear in a longer prefix, and the numbers in the length-$n$ prefix that are not in $[1..n]$ are in $[1..m]$ for some $m > n$. Since the terms eventually go to zero, the sum of any $2k$ terms with eventually increasing indices must also go to zero.

This condition is far from necessary, because the following permutation is sum-preserving but has $\limsup_{n\to\infty} \frac{f(n)}{n} = \infty$ and $\liminf_{n\to\infty} \frac{f(n)}{n} = 0$:

$1;3,2;9,8,7,6,5,4;\cdots$

where the $k$-th block has $k!$ elements and is simply reversed. The reason is that for any $ε > 0$ there is some natural $k$ such that all the partial sums of any block beyond the $k$-th have magnitude less than $ε$, and hence also the partial sums of the block's reverse. This means that the rearranged series has the same sum by Cauchy convergence.