In this nice answer to the question "Prove that $a_{n}=0$ for all $n$, if $\sum a_{kn}=0$ for all $k\geq 1$" it is required to show that following sum converges to $a_1$: $$ \sum_{m=1}^\infty \left( \mu(m)\sum_{n=1}^\infty a_{mn}\right), $$ where $mn=m\times n$. This is not true in general when $\sum_n a_{mn}$ is not identically zero, but we are focusing on the case $\sum_n a_{mn}=0.$
Of course, the following cutoff version is always true: $$ a_1= \sum_{mn\leq N} \mu(m)a_{mn}. $$ However, it is really difficult to show that $$ \sum_{mn\leq N} a_{mn} \to \sum_{m=1}^\infty \left( \mu(m)\sum_{n=1}^\infty a_{mn}\right) $$ as $N \to\infty$. I find myself always in need of some bounds on $\sum a_n$ related to the number of devisors of $n$, which is not provided by the fact that $\sum a_n$ is abosolutely convergent.
Absolute convergence of the double series is of course not always achievable.
And the entire thing is a delicate unstable equilibrium, because if we just require $|\sum \mu(m) a_{mn}| < \epsilon$ for all $n$, then it appears that we in general cannot conclude that $a_1<C\epsilon$ for some fixed constant $C$: there are counterexamples. We want $|\sum \mu(m) a_{mn}|$ to decrease fast with $m$, but unfortunately, we can make it decrease arbitrarily slow, by choosing $a_n$ to be a sequence with many zeros. To put it shortly, small perturbation in the sense of something like $l^\infty$ norm is unstable, and we are not free to choose other stronger norms.
So, how to complete the proof of convergence?
It would be nice if further results of stability can be discussed as well.