We first prove that $S$ is a subset of $R$. Let $f$ be a member of $\mathcal{F}$ and write $P(x, y)$ for $f(x + f(y)) = f(x) + f(y)$.
Begin by showing that $1 + 1/n$ belongs to $R$ for all positive integers $n$. To this end, induct on $n$ to prove that $f(nf(0)) = (n + 1)f(0)$; letting $z = nf(0)$, then $f(z) = (1 + 1/n)z$, so $1 + 1/n$ is a member of $R$. The base case, $n = 1$, is provided by $P(0, 0)$.

For the induction step, assume $f(nf(0)) = (n+1)f(0)$ and consider $P(0, nf(0))$ to write
$$\begin{aligned} f((n+1)f(0)) &= f(f(nf(0))) = f(0 + f(nf(0))) = f(0) + f(nf(0)) \\ &= f(0) + (n+1)f(0) = (n+2)f(0), \end{aligned}$$
as desired. This completes the induction.

To prove that $1-1/n$ belongs to $R$ for all positive integers $n$, notice that $P(-f(0), 0)$ yields $f(-f(0)) = 0$, and an inductive argument along the lines above then shows that $f(-nf(0)) = (-n+1)f(0)$. Letting $z = -nf(0)$, then $f(z) = (1 - 1/n)z$, so $1 - 1/n$ is a member of $R$. Consequently, $S$ is a subset of $R$.

Let $r \neq 1$ and consider the function $g: [0, 1) \to \mathbb{R}$,
$$g(x) = \begin{cases} 1 & \text{if } rx/(1-r) \text{ is an integer} \\ 0 & \text{otherwise.} \end{cases}$$