We claim the answer is $r = 1 - \frac{1}{4k}$ for all positive integers $k$.

Suppose $A$ and $B$ are points in $S$. Place $A$ and $B$ on the complex plane such that $A = 0$ and $B = 1$, and let $\omega = \cos\theta + i\sin\theta = r + i\sqrt{1-r^2}$. We claim that any good set containing $A$ and $B$ contains the good set $T$ of numbers of the form $f(\omega)$, where $f(x)$ is a Laurent polynomial—that is, $x^n f(x)$ is a polynomial for some (possibly negative) integer $n$—with integer coefficients, and $f(1) = 0$ or $1$.

We first show that $T$ is good. Indeed, for any $p \in T$, the rotation $R_p$ about $p$ by $\theta$ sends $z \mapsto R_p(z) = (z-p)\omega + p$. If $p = f(\omega)$ and $z = g(\omega) \in T$, then $h(x) = (g(x)-f(x)) \cdot x + f(x)$ satisfies $R_p(z) = h(\omega)$ and $h(1) = g(1) = 0$ or $1$, so $R_p(z) \in T$. Similarly, $R_p^{-1}(z) = (z-p) \cdot \omega^{-1} + p \in T$, so $T$ is good.

Next we show that any good set $S$ containing $0$ and $1$ contains all of $T$. Let $f(x)$ be any Laurent polynomial with integer coefficients with $f(1) = 0$ or $1$. We will show that $f(\omega) \in S$. We induct on the sum $s$ of the absolute value of the coefficients of $f(x)$. The case $s=0$ is trivial. For $s=1$, $f(\omega)$ must be a power of $\omega$, which is of the form $R_0^k(1) \in S$. For $s \ge 2$, $f(x)$ must have at least one positive and one negative coefficient, so write $f(x) = x^a - x^b + g(x)$, where the sum of the absolute values of the coefficients of $g(x)$ is $s-2$. Since $g(1) = f(1)$, $g(\omega) \in S$ by induction. Then
$$x^{-a}f(x) = (x^{-b}g(x) - 1) \cdot x^{b-a} + 1,$$
so $f(\omega) = R_0^a \circ R_1^{b-a} \circ R_0^{-b}(g(\omega)) \in S$, completing the induction.

It follows that $r$ satisfies the desired condition if and only if we can write $\frac{1}{2} = f(\omega) \in T$. Then there is a polynomial $g(x) = x^n(2f(x)-1)$, all but one of whose coefficients are even, that has $\omega$ as a root. If $r = \pm 1$, so $\omega = \pm 1$, this is clearly impossible. Otherwise, let $r = \frac{a}{b}$ with $a$ and $b$ relatively prime. Then the minimal polynomial $p(x)$ of $\omega$ over the integers is $bx^2 - 2ax + b$ if $b$ is odd and $\frac{b}{2}x^2 - ax + \frac{b}{2}$ if $b$ is even. This minimal polynomial must divide $g(x)$. But since $g(x)$ is congruent to a power of $x$ modulo 2, $p(x)$ must also be congruent to a power of $x$ modulo 2. The only possibility is for $b$ to be a multiple of 4 (and $a$ to be odd).

Finally, since $g(1) = 2f(1) - 1 = \pm 1$ and $p(1) | g(1)$, we must have that $p(1) = \pm 1$ as well. Since $p(1) = \frac{b}{2} - a + \frac{b}{2} = b - a$, we must have $b = a \pm 1$. Since $|r| \le 1$, $|a| \le |b|$. We find then that the only possibilities are $r = 1 - \frac{1}{4k}$, where we can let $a = 4k-1$ and $b = 4k$, so $p(1) = 1$. Then letting $f(x) = \frac{1}{2}(x^{-1}p(x) + 1) = kx - (2k-1) + kx^{-1}$, we find that $f(1) = 1$, so $\frac{1}{2} = f(\omega) \in T$, as desired.