The answer is that $r$ has this property if and only if $r=\frac{4 n-1}{4 n}$ for some integer $n$.
Throughout the solution, we will let $r=\frac{a}{b}$ with $b>0$ and $\operatorname{gcd}(a, b)=1$. We also let
$$
\omega=e^{i \theta}=\frac{a}{b} \pm \frac{\sqrt{b^{2}-a^{2}}}{b} i .
$$

This means we may work with complex multiplication in the usual way; the rotation of $z$ through center $c$ is given by $z \mapsto \omega(z-c)+c$.
For most of our proof, we start by constructing a good set as follows.
- Start by letting $S_{0}=\{0,1\}$.
- Let $S_{i}$ consist of $S_{i-1}$ plus all points that can be obtained by rotating a point of $S_{i-1}$ through a different point of $S_{i-1}$ (with scale factor $\omega$ ).
- Let $S_{\infty}=\bigcup_{i \geq 0} S_{i}$.

The set $S_{\infty}$ is the (minimal, by inclusion) good set containing 0 and 1 . We are going to show that for most values of $r$, we have $\frac{1}{2} \notin S_{\infty}$.
Claim - If $b$ is odd, then $\frac{1}{2} \notin S_{\infty}$.

Proof. Idea: denominators that appear are always odd.
Consider the ring
$$
A=\mathbb{Z}_{\{b\}}=\left\{\frac{s}{t}|s, t \in \mathbb{Z}, t| b^{\infty}\right\}
$$
which consists of all rational numbers whose denominators divide $b^{\infty}$. Then, $0,1, \omega \in$ $A\left[\sqrt{b^{2}-a^{2}}\right]$ and hence $S_{\infty} \subseteq A\left[\sqrt{b^{2}-a^{2}}\right]$ too. (This works even if $\sqrt{b^{2}-a^{2}} \in \mathbb{Z}$, in which case $S_{\infty} \subseteq A=A\left[\sqrt{b^{2}-a^{2}}\right]$.)
But $\frac{1}{2} \notin A\left[\sqrt{b^{2}-a^{2}}\right]$.
Claim - If $b$ is even and $|b-a| \neq 1$, then $\frac{1}{2} \notin S_{\infty}$.

Proof. Idea: take modulo a prime dividing $b-a$.
Let $D=b^{2}-a^{2} \equiv 3(\bmod 4)$. Let $p$ be a prime divisor of $b-a$. Because $\operatorname{gcd}(a, b)=1$, we have $p \neq 2$ and $p \nmid b$.
Consider the ring
$$
A=\mathbb{Z}_{(p)}=\left\{\left.\frac{s}{t} \right\rvert\, s, t \in \mathbb{Z}, p \perp t\right\}
$$
which consists of all rational numbers whose denominators are coprime to $p$. Then, $0,1, \omega \in A[\sqrt{-D}]$ and hence $S_{\infty} \subseteq A[\sqrt{-D}]$ too.
Now, there is a well-defined "mod-p" ring homomorphism
$$
\Psi: A[\sqrt{-D}] \rightarrow \mathbb{F}_{p} \quad \text { by } \quad x+y \sqrt{-D} \mapsto x \bmod p
$$
which commutes with addition and multiplication (as $p \mid D$ ). Under this map,
$$
\omega \mapsto \frac{a}{b} \bmod p=1 .
$$

Consequently, the rotation $z \mapsto \omega(z-c)+c$ is just the identity map modulo $p$. In other words, the pre-image of any point in $S_{\infty}$ under $\Psi$ must be either $\Psi(0)=0$ or $\Psi(1)=1$. However, $\Psi(1 / 2)=1 / 2$ is neither of these. So this point cannot be achieved.
Claim - Suppose $a=2 n-1$ and $b=2 n$ for $n$ an odd integer. Then $\frac{1}{2} \notin S_{\infty}$

Proof. Idea: $\omega$ is "algebraic integer" sans odd denominators.
This time, we define the ring
$$
B=\mathbb{Z}_{(2)}=\left\{\left.\frac{s}{t} \right\rvert\, s, t \in \mathbb{Z}, t \text { odd }\right\}
$$
of rational numbers with odd denominator. We carefully consider the ring $B[\omega]$ where
$$
\omega=\frac{2 n-1 \pm \sqrt{1-4 n}}{2 n} .
$$

So $S_{\infty} \subseteq B[\omega]$ as $0,1, \omega \in B[\omega]$.
I claim that $B[\omega]$ is an integral extension of $B$; equivalently that $\omega$ is integral over $B$. Indeed, $\omega$ is the root of the monic polynomial
$$
(T-1)^{2}+\frac{1}{n}(T-1)-\frac{1}{n}=0
$$
where $\frac{1}{n} \in B$ makes sense as $n$ is odd.
On the other hand, $\frac{1}{2}$ is not integral over $B$ so it is not an element of $B[\omega]$.
It remains to show that if $r=\frac{4 n-1}{4 n}$, then goods sets satisfy the midpoint property. Again starting from the points $z_{0}=0, z_{1}=1$ construct the sequence
$$
\begin{array}{l}
z_{2}=\omega\left(z_{1}-z_{0}\right)+z_{0} \\
z_{3}=\omega^{-1}\left(z_{0}-z_{2}\right)+z_{2} \\
z_{4}=\omega^{-1}\left(z_{2}-z_{3}\right)+z_{3} \\
z_{5}=\omega\left(z_{3}-z_{4}\right)+z_{4}
\end{array}
$$
as shown in the diagram below.
This construction shows that if we have the length-one segment $\{0,1\}$ then we can construct the length-one segment $\{2 r-2,2 r-1\}$. In other words, we can shift the segment to the left by
$$
1-(2 r-1)=2(1-r)=\frac{1}{2 n} \text {. }
$$

Repeating this construction $n$ times gives the desired midpoint $\frac{1}{2}$.
Let $r$ be a rational number in the interval $[-1,1]$ and let $\theta=\cos ^{-1} r$. Call a subset $S$ of the plane good if $S$ is unchanged upon rotation by $\theta$ around any point of $S$ (in both clockwise and counterclockwise directions). Determine all values of $r$ satisfying the following property: The midpoint of any two points in a good set also lies in the set.