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bump bump!

We can bound this sum to be a finite amount above \[\sum_{x\in S}\frac 1x.\]This new sum is less than \[ \sum_{b\ge 2}\sum_{n\ge 2}\frac{1}{b^n} =\sum_{b\ge 2}\frac{1}{b^2-b}=1. \]Therefore, the sum converges. Let $T=\mathbb{N}-S-\{1\}$. Then note that \begin{align*} & \sum_{x\in S}\frac{1}{x-1} \\ =& \sum_{x\in S}\sum_{k\ge 1}x^{-k} \\ =& \sum_{x\in T}\sum_{m\ge 2}\sum_{k\ge 1}x^{-mk} \\ =& \sum_{x\ge 2}\sum_{m\ge 2} x^{-k} \\ =& \sum_{x\ge 2}\frac{1}{x^2-x} \\ =& \boxed{1}. \end{align*}