Re-notate $\sigma$ with $f: [n] \to [n]$, a bijection, and let $s = k + \ell$. Clearly we must have $f(1) = 1+k$ and $f^{-1}(1) = 1+\ell$. For $i=2, 3, \dots, \ell$ we must have $f(i) = i+k$, and for $i=2, 3, \dots, k$ we must have $f^{-1}(i)=i+\ell$.
This determines $f$ on $\{1, 2, \dots, s\} \to \{1, 2, \dots, s\}$, and furthermore it forces $f(1 + s) = (1+s) + k$ and $f^{-1}(1+s) = (1+s) + \ell$. We may repeat our reasoning to determine $f$ on $\{s+1, s+2, \dots, 2s\}$, etc. When we have finished, we have a total of $n = ms$ for some $m \in \mathbb{N}$, as desired.