For a conditionally convergent series $\sum a_n$ over $\Bbb C$, let
$M\subset \Bbb N$ be a set such that the sub-series $\displaystyle \sum_{n\in M} a_n$ converges absolutely.
$\sigma:\Bbb N \to \Bbb N$ be a permutation such that $\sigma$ restricted to $\Bbb N\setminus M$ is the identity, i.e. $\sigma$ only reorders elements of $(a_n)$ for $n\in M$.
Then: $$\sum a_{\sigma(n)}~\text{ converges with the same value like }\ \sum a_n\tag1$$
I don't actually know if $(1)$ does hold, so how would a proof look like, resp. how would a counter-example go? Is there some intuition that tells which route to take?
Your claim does hold.
Let $M=\{m_1,m_2,m_3,...\}$ be an increasing enumeration. Denote by $S_n=\sum_{k=1}^na_n$ and $S^\sigma_n=\sum_{k=1}^na_{\sigma(n)}$ the partial sums, we have
$$|S^\sigma_n-S_n|\leq\sum_{k\in I_n}|a_{\sigma(m_k)}|+|a_{m_k}|$$
where $I_n=\{k\in\mathbb N:m_k\leq n\ \text{and}\ \sigma^{-1}(m_k)>n\}$. To see why the above bound holds, note that if both $m_k$ and $\sigma^{-1}(m_k)$ are bounded by $n$, the term $a_{m_k}$ will appear in both $S_n$ and $S^\sigma_n$ and therefore be cancelled. Thus, the terms that are not cancelled are $a_{m_k}$ and $a_{\sigma(m_k)}$ with $k\in I_n$. Since $I_n$ is finite, it has a minimum and we have
$$|S^\sigma_n-S_n|\leq2\sum_{k\geq\min(I_n)}|a_{m_k}|.$$
As $n\to\infty$, we must have $\min(I_n)\to\infty$ because $\sigma^{-1}(n)$ is finite for each $n$, this implies $\sum_{k\geq\min(I_n)}|a_{m_k}|\to0$ since $a_k$ is absolutely summable over $M$. Hence $|S^\sigma_n-S_n|\to0$ and your claim has been proved.
(The convergence is only pointwise in $\sigma$)