If $\mathbb{N}$ is the set of positive integers, then it is formally false, we may add arbitrary number of zero terms to our series. But if all $a_i$ and $b_i$ are different from 0 complex numbers, and the series converge absolutely, then the answer is yes. Just rearrange both sequences so that $|a_1|\geq |a_2|\geq |a_3|\geq \dots$ and the same for $b_i$'s. Note that $\limsup |\sum a_i^s|/|a_1|^s$ equals $\max\{k: |a_k|=a_1\}$. It follows that $|b_1|=|a_1|$ and both sequences have the same number of terms with absolute value equal to $|a_1|$. Denote the number of such terms by $k$, let them be $a_1,\dots,a_k$ and $b_1,b_2,\dots,b_k$ respectively. I claim that $A:=\{a_1,\dots,a_k\}=\{b_1,\dots,b_k\}=:B$ in the sense of multi-set equalities. Indeed, without loss of generality we may suppose that $|a_1|=1$. Choose a sequence $\{s_i\}$ such that $a_m^{s_i}$ and $b_m^{s_i}$ tends to $1$ for all $m\in \{1,2,\dots,k\}$. Such sequence does exist by Kronecker lemma. Then for fixed postive integer $k$ one has $\lim_j \sum_i a_i^{s_j+p}=\sum_{i=1}^k a_i^p$, the same for $b$'s. That is, power sums of arrays $A$ and $B$ are equal, it follows that those arrays are equal. Then remove such arrays from both sequences and proceed the same way.

For real sequences, from convergence of $\sum a_i^2$ the absolute convergence of $\sum a_i^k$ for $k>2$ follows, and if we consider only $k$'s divisible by 3 we conclude that $\{a_i^3\}$ is a permutation of $\{b_i^3\}$, what desired.

I do not know the answer for conditionally convergent complex sequences.