Solution 1. Argue indirectly. Agree, as usual, that the empty sum is $0$ to consider rationals in $[0,1)$; adjoining $0$ causes no harm, since $\sum_{x \in F} 1 / x = 0$ for no nonempty finite subset $F$ of $S$. For every rational $r$ in $[0,1)$, let $F_{r}$ be the unique finite subset of $S$ such that $\sum_{x \in F_{r}} 1 / x = r$. The argument hinges on the lemma below.

Lemma. If $x$ is a member of $S$ and $q$ and $r$ are rationals in $[0,1)$ such that $q - r = 1 / x$, then $x$ is a member of $F_{q}$ if and only if it is not one of $F_{r}$.

Proof. If $x$ is a member of $F_{q}$, then
$$\sum_{y \in F_{q} \setminus \{x\}} \frac{1}{y} = \sum_{y \in F_{q}} \frac{1}{y} - \frac{1}{x} = q - \frac{1}{x} = r = \sum_{y \in F_{r}} \frac{1}{y}$$
so $F_{r} = F_{q} \setminus \{x\}$, and $x$ is not a member of $F_{r}$. Conversely, if $x$ is not a member of $F_{r}$, then
$$\sum_{y \in F_{r} \cup \{x\}} \frac{1}{y} = \sum_{y \in F_{r}} \frac{1}{y} + \frac{1}{x} = r + \frac{1}{x} = q = \sum_{y \in F_{q}} \frac{1}{y}$$
so $F_{q} = F_{r} \cup \{x\}$, and $x$ is a member of $F_{q}$.

Consider now an element $x$ of $S$ and a positive rational $r < 1$. Let $n = \lfloor r x \rfloor$ and consider the sets $F_{r - k / x}, k = 0, \ldots, n$. Since $0 \leqslant r - n / x < 1 / x$, the set $F_{r - n / x}$ does not contain $x$, and a repeated application of the lemma shows that the $F_{r - (n - 2k) / x}$ do not contain $x$, whereas the $F_{r - (n - 2k - 1) / x}$ do. Consequently, $x$ is a member of $F_{r}$ if and only if $n$ is odd.

Solution 2. A finite $S$ clearly satisfies (2), so let $S$ be infinite. If $S$ fails both conditions, so does $S \setminus \{1\}$. We may and will therefore assume that $S$ consists of integers greater than $1$. Label the elements of $S$ increasingly $x_{1} < x_{2} < \cdots$, where $x_{1} \geqslant 2$.

We first show that $S$ satisfies (2) if $x_{n+1} \geqslant 2 x_{n}$ for all $n$. In this case, $x_{n} \geqslant 2^{n-1} x_{1}$ for all $n$, so
$$s = \sum_{n \geqslant 1} \frac{1}{x_{n}} \leqslant \sum_{n \geqslant 1} \frac{1}{2^{n-1} x_{1}} = \frac{2}{x_{1}}$$
If $x_{1} \geqslant 3$, or $x_{1} = 2$ and $x_{n+1} > 2 x_{n}$ for some $n$, then $\sum_{x \in F} 1 / x < s < 1$ for every finite subset $F$ of $S$, so $S$ satisfies (2); and if $x_{1} = 2$ and $x_{n+1} = 2 x_{n}$ for all $n$, that is, $x_{n} = 2^{n}$ for all $n$, then every finite subset $F$ of $S$ consists of powers of $2$, so $\sum_{x \in F} 1 / x \neq 1 / 3$ and again $S$ satisfies (2).

Finally, we deal with the case where $x_{n+1} < 2 x_{n}$ for some $n$. Consider the positive rational $r = 1 / x_{n} - 1 / x_{n+1} < 1 / x_{n+1}$. If $r = \sum_{x \in F} 1 / x$ for no finite subset $F$ of $S$, then $S$ satisfies (2).

We now assume that $r = \sum_{x \in F_{0}} 1 / x$ for some finite subset $F_{0}$ of $S$, and show that $S$ satisfies (1). Since $\sum_{x \in F_{0}} 1 / x = r < 1 / x_{n+1}$, it follows that $x_{n+1}$ is not a member of $F_{0}$, so
$$\sum_{x \in F_{0} \cup \{x_{n+1}\}} \frac{1}{x} = \sum_{x \in F_{0}} \frac{1}{x} + \frac{1}{x_{n+1}} = r + \frac{1}{x_{n+1}} = \frac{1}{x_{n}}.$$
Consequently, $F = F_{0} \cup \{x_{n+1}\}$ and $G = \{x_{n}\}$ are distinct finite subsets of $S$ such that $\sum_{x \in F} 1 / x = \sum_{x \in G} 1 / x$, and $S$ satisfies (1).