Solution:

No such set exists, which we will show by a proof by contradiction. Suppose $S$ has the stated property. Clearly $M$ is then infinite, and we may assume without loss of generality that $1 \notin M$. We denote the elements of $M$, ordered by size, by $m_{1} < m_{2} < m_{3} < \ldots$.

Case 1: We have $m_{i} \geq 2 m_{i-1}$ for all $i \geq 2$. Then $m_{i} \geq 2^{i-1} m_{1}$ holds for all $i \geq 1$, and it follows that
$$r^{*} := \sum_{i=1}^{\infty} \frac{1}{m_{i}} \leq \sum_{i=0}^{\infty} \frac{1}{2^{i} m_{1}} = \frac{2}{m_{1}},$$
that is, if $m_{1} \geq 3$ or $m_{i} > 2 m_{i-1}$ holds for at least one $i$, then $r^{*} < 1$. In this case, however, $M$ does not satisfy the condition stated in the problem text, contrary to the assumption, if $s \in (r^{*}, 1)$ is chosen. Thus we must have $m_{1} = 2$ and $m_{i} = 2 m_{i-1}$ for all $i$, so $M$ consists exactly of the powers of two greater than $1$. Since $1 / 3 = \sum_{i=2}^{\infty} 2^{-i}$ cannot be written as a finite sum of reciprocals of powers of two, we also get a contradiction in this case.

Case 2: There exists an $i > 1$ such that $m_{i} < 2 m_{i-1}$. We consider
$$r := \frac{1}{m_{i-1}} - \frac{1}{m_{i}} < \frac{1}{m_{i}}.$$
By assumption there exists a finite subset $S \subset M$ such that $\sum_{s \in S} 1 / s = r$ holds. Because $s < 1 / m_{i}$, we have $m_{i} \notin S$. But then $S_{1} = S \cup \{m_{i}\}$ and $S_{2} = \{m_{i-1}\}$ are two distinct finite subsets of $M$ such that the sums of the reciprocals of $S_{1}$ and $S_{2}$ both equal $1 / m_{i-1}$, which is a contradiction.