I think so. I've re-written the whole thing to make it a bit cleaner and integrate the last two posts. The proof of convergence of the absolute double sum is done at the start. I hope it makes sense now.
Theorem
If the sequence \sum_{n=1}^\infty a_n is absolutely convergent with limit L, and {\pi:\mathbb N\times\mathbb N\to\mathbb N} is a bijection, then L'\equiv\sum_{k=1}^\infty\sum_{l=1}^\infty a_{\pi(k,l)} is a convergent sum, which is equal to ##L##.
Proof
First we prove that \sum_{k=1}^\infty\sum_{l=1}^\infty|a_{\pi(k,l)}| exists.
If it does not exist then either one or more of the inner sums is divergent, or the inner sums converge but the outer sum diverges.
First assume that the inner sum for k diverges. Then for any R>0 there exists G\in\mathbb N such that \sum_{l=1}^G|a_{\pi(k,l)}|>R. Then \sum_{n=1}^N|a_n|>R where
$$N\equiv\max_{l\in\{1,...,G\}}\pi(k,l)$$
which contradicts the absolute convergence of \sum_{n=0}^\infty a_n. So no inner sum can diverge.
Next assume the outer sum diverges. Then for any R>0 we can find G\in\mathbb N such that
$$\sum_{k=1}^G\sum_{l=1}^\infty|a_{\pi(k,l)}|>R+1$$
Since there is a finite number of inner sums and they all converge, we can find F\in\mathbb N such that for all l\in\{1,...,F\} we have \sum_{l=F+1}^\infty|a_{\pi(k,l)}|<\frac1G.
Then we have
$$\sum_{k=1}^G\sum_{l=1}^F|a_{\pi(k,l)}|
=\sum_{k=1}^G\sum_{l=1}^\infty|a_{\pi(k,l)}|-
\sum_{k=1}^G\sum_{l=F+1}^\infty|a_{\pi(k,l)}|
>R+1+G\cdot\frac1G=R$$
Then \sum_{n=1}^N|a_n|>R where
$$N\equiv\max_{\substack{k\in\{1,...,G\}\\l\in\{1,...,F\}}}\pi(k,l)$$
which contradicts the absolute convergence of \sum_{n=0}^\infty a_n. So the outer sum cannot diverge either. Hence the double sum of absolute values converges.
We see then that the sum \sum_{k=1}^\infty\sum_{l=1}^\infty a_{\pi(k,l)} must also converge to a limit L' because sums of absolute values are no less than sums of signed numbers, so if a signed sum diverges then the sum formed by applying the same summation operators to the absolute values of the summands must also diverge.
Now we try to prove the result.
First, given \epsilon>0, we choose N\in\mathbb N such that \sum_{n=N+1}^\infty|a_n|<\frac\epsilon9, which we can do by the absolute convergence of that series.
Next we set
$$H\equiv\max_{k\in\{1,2,...,N\}}\left(\pi^{-1}(k)\right)_1$$
where the 1 subscript indicates the first component of an element of \mathbb N\times\mathbb N.
Next we choose G\in\mathbb N such that \sum_{k=G+1}^\infty\sum_{l=1}^\infty|a_{\pi(k,l)}|<\frac\epsilon9, which we can do by the convergence of the double sum, proven above.
Then define M\equiv\max(G,H).
Next, for each k\in\{1,...,M\} we choose T'(k) such that {\sum_{l=T'(k)+1}^\infty|a_{\pi(k,l)}|<\frac\epsilon{9M}}, which again we can do by convergence of the double sum above. We then set T\equiv\max_{k\in\{1,...,M\}}T'(k).
Armed with these definitions, we proceed as follows:
\begin{align}
|L'-L|&\equiv\left|\sum_{k=1}^\infty\sum_{l=1}^\infty a_{\pi(k,l)}-\sum_{n=1}^\infty a_n\right|\\
&=\left|\left(\sum_{k=1}^M\sum_{l=1}^T
+\sum_{k=1}^M\sum_{l=T+1}^\infty
+\sum_{k=M+1}^\infty\sum_{l=1}^\infty
\right)a_{\pi(k,l)}
-\left(\sum_{n=1}^N+\sum_{n=N+1}^\infty\right) a_n\right|\\
&\leq\left|\sum_{k=1}^M\sum_{l=1}^T a_{\pi(k,l)}-\sum_{n=1}^N a_n\right|
+\left|\sum_{k=1}^M\sum_{l=T+1}^\infty a_{\pi(k,l)}\right|
+\left|\sum_{k=M+1}^\infty\sum_{l=1}^\infty a_{\pi(k,l)}\right|
+\left|\sum_{n=N+1}^\infty a_n\right|\\
&<\left|\sum_{k=1}^M\sum_{\substack{l=1\\\pi(k,l)>n}}^T a_{\pi(k,l)}\right|
+\sum_{k=1}^M\sum_{l=T+1}^\infty \left|a_{\pi(k,l)}\right|
+\sum_{k=M+1}^\infty\sum_{l=1}^\infty \left|a_{\pi(k,l)}\right|
+\sum_{n=N+1}^\infty \left|a_n\right|\\
&<\sum_{k=1}^M\sum_{\substack{l=1\\\pi(k,l)>N}}^T \left|a_{\pi(k,l)}\right|
+\sum_{k=1}^M\sum_{l=T'(k)+1}^\infty \left|a_{\pi(k,l)}\right|
+\sum_{k=G+1}^\infty\sum_{l=1}^\infty \left|a_{\pi(k,l)}\right|
+\frac\epsilon9\\
&<\sum_{n=N+1}^\infty|a_n|
+\sum_{k=1}^\infty\frac\epsilon{9M}+\frac\epsilon9
+\frac\epsilon9\\
&=\frac{\epsilon}9
+\frac{\epsilon}3=\frac{4\epsilon}9<\epsilon\\
\end{align}
Since |L'-L|<\epsilon for all positive \epsilon, it must be zero.
Phew!