It is well-known that if $\sum_{n=1}^\infty a_n$ is an absolutely convergent complex series and $\mathbb N$ is partitioned as $J_1,J_2,\dots$, then the series $\sum_{j\in J_n}a_j$ for all $n$ and $\sum_{n=1}^\infty\sum_{j\in J_n}a_j$ are both absolutely convergent, and $\sum_{n=1}^\infty a_n=\sum_{n=1}^\infty\sum_{j\in J_n}a_j$.
I am trying to make sense of the same identity for infinite products. The convention I'm using is the following
The product $\prod_{n=1}^\infty z_n$ is called convergent if there exists an $n_0$ such that the partial products $\prod_{j=n_0}^nz_j$ converge to a non-zero number as $n\to\infty$. Write $z_n=1+w_n$, the product is said to converge absolutely if $\prod_n(1+|w_n|)$ converges. It is a theorem that an absolutely convergent product is convergent and that $\prod_n(1+|w_n|)$ converges if and only if $\sum_n|w_n|$ does.
Suppose $\prod_{n=1}^\infty z_n$ converges absolutely and $\Bbb N=\bigcup_{n=1}^\infty J_n$ is a partition, I want to show that $$\prod_{n=1}^\infty z_n=\prod_{n=1}^\infty\prod_{j\in J_n}z_j,$$ with the RHS converging absolutely. By the characterisation of absolute convergence, we know that $\sum_{n=1}^\infty|z_n-1|$ converges, so the series $\sum_{j\in J_n}|z_j-1|$ converge for all $n$ and hence all the products $\prod_{j\in J_n}z_j$ converge absolutely. What I'm having trouble with is showing $\prod_{n=1}^\infty\prod_{j\in J_n}z_j$ converges (absolutely or not) and showing it is equal to the original product. By the characterisation of absolute convergence, it would be enough to show that $\sum_{n=1}^\infty|(\prod_{j\in J_n}z_j)-1|$ converges, but this has no discernible relation to the series $\sum_{n=1}^\infty\sum_{j\in J_n}|z_j-1|$, which we do know converges by the "associativity" for absolutely convergent series.