Solution:
Answer: $\{-1,1\}$

The key step is to notice that for such a function $f$, $f^{\prime}(x) \neq 0$ for any $x$.

Assume, for sake of contradiction, that there exists $0 < y < 1$ such that $f^{\prime}(y) = 0$. Since $f^{\prime}$ is a continuous function, there is some small interval $(c, d)$ containing $y$ such that $|f^{\prime}(x)| \leq \frac{1}{2}$ for all $x \in (c, d)$. Now there exists some $n, a$ such that $\frac{a}{2^{n}}, \frac{a+1}{2^{n}}$ are both in the interval $(c, d)$. From the definition,
$$\frac{f\left(\frac{a+1}{2^{n}}\right) - f\left(\frac{a}{2^{n}}\right)}{\frac{a+1}{2^{n}} - \frac{a}{2^{n}}} = 2^{n}\left(\frac{b'}{2^{n}} - \frac{b}{2^{n}}\right) = b' - b$$
where $b, b'$ are integers; one is odd, and one is even. So $b' - b$ is an odd integer. Since $f$ is differentiable, by the mean value theorem there exists a point where $f^{\prime} = b' - b$. But this point is in the interval $(c, d)$, and $|b' - b| > \frac{1}{2}$. This contradicts the assumption that $|f^{\prime}(x)| \leq \frac{1}{2}$ for all $x \in (c, d)$.

Since $f^{\prime}(x) \neq 0$, and $f^{\prime}$ is a continuous function, $f^{\prime}$ is either always positive or always negative. So $f$ is either increasing or decreasing. $f\left(\frac{1}{2}\right) = \frac{1}{2}$ always. If $f$ is increasing, it follows that $f\left(\frac{1}{4}\right) = \frac{1}{4}$, $f\left(\frac{3}{4}\right) = \frac{3}{4}$, and we can show by induction that indeed $f\left(\frac{a}{2^{n}}\right) = \frac{a}{2^{n}}$ for all integers $a, n$. Since numbers of this form are dense in the interval $(0,1)$, and $f$ is a continuous function, $f(x) = x$ for all $x$.

It can be similarly shown that if $f$ is decreasing, $f(x) = 1 - x$ for all $x$. So the only possible values of $f^{\prime}\left(\frac{1}{2}\right)$ are $-1, 1$.