Define $S_{k,n}$ like in this question. I think I have proved the identity $$\sum_{m\in S_{k,n+1}}m^{-1}=\sum_{m\in S_{k,n}}m^{-1}+\frac1{p_{n+1}}\sum_{m\in S_{k-1,n}}m^{-1} $$for all $n\ge k\ge1$. In other words, for these $k,n$, the sets satisfy $S_{k,n+1}=S_{k,n} \ \cup (p_{n+1}) \cdot S_{k-1,n}$ where $(p_{n+1}) \cdot S_{k-1,n}$ stands for $\{p_{n+1}\cdot m \ | m\in S_{k-1,n}\}.$
I am going to write down my proof below; I would like to have feedback on it. Besides that, the $S_{k,n}$ have to do with primes, but the identity (and the proof) seems to hold for any family of sets defined in the same way, so as an aside I was wondering if there is a combinatorial approach.
Let $p_i$ be the $i$-th prime number. If $k=1$, the equality reduces to $\sum_{i=1}^{n+1}p_i^{-1}=\sum_{i=1}^{n}p_i^{-1}+(p_{n+1})^{-1}.$ For larger $k$ we just need to expand the left-hand side, abiding by the convention to view any occurring empty product as $1$: \begin{align}\sum_{m\in S_{k,n+1}}m^{-1}&=\frac1{p_{k-1}\prod_{i=1}^{k-2}p_i}\sum_{j=k}^{n+1}\frac1{p_j}+\frac1{p_k\prod_{i=1}^{k-2}p_i}\sum_{j=k+1}^{n+1}\frac1{p_i}+\cdots+\frac1{p_np_{n+1}\prod_{i=1}^{k-2}p_i}\\&+\frac1{p_{k}\prod_{i=2}^{k-1}p_i}\sum_{j=k+1}^{n+1}\frac1{p_j}+\frac1{p_{k+1}\prod_{i=2}^{k-1}p_i}\sum_{j=k+2}^{n+1}\frac1{p_i}+\cdots+\frac1{p_np_{n+1}\prod_{i=2}^{k-1}p_i}\\&\vdots\\&+\frac1{p_{n-1}\prod_{i=n-k+1}^{n-2}p_i}\left(\frac1{p_n}+\frac1{p_{n+1}}\right)+\frac1{p_{n}p_{n+1}\prod_{i=n-k+1}^{n-2}p_i}\\&=\sum_{m\in S_{k,n}}m^{-1}+\frac1{p_{n+1}}\left(\frac1{p_{k-1}\prod_{i=1}^{k-2}p_i}+\frac1{p_k\prod_{i=1}^{k-2}p_i}+\cdots+\frac1{p_n\prod_{i=1}^{k-2}p_i}\right)\\&+\frac1{p_{n+1}}\left(\frac1{p_{k}\prod_{i=1}^{k-1}p_i}+\frac1{p_{k+1}\prod_{i=1}^{k-1}p_i}+\cdots+\frac1{p_n\prod_{i=1}^{k-1}p_i}\right)\\&\vdots\\&+\frac1{p_{n+1}}\left(\frac1{p_{n-1}\prod_{i=n-k+1}^{n-2}p_i}+\frac1{p_{n}\prod_{i=n-k+1}^{n-2}p_i}\right)\\&=\sum_{m\in S_{k,n}}m^{-1}+\frac1{p_{n+1}}\sum_{m\in S_{k-1,n}}m^{-1}. \end{align}