Suppose that $a,b \in \mathbb C \backslash \{0\}, a \neq 1$. Prove that there is some $R > 0$ such that there are no holomorphic functions $f: \{z: |z| < R \} \to \mathbb C \backslash \{0,1\}$ with $f(0) = a$ and $f'(0)= b$.
Suppose on the contrary that for all $R > 0$ there is at least one holomrphic function that satisfy the conditions. Then for any $n \in \mathbb N$, we would have increasing functions $f_n: \{z: |z| < n \} \to \mathbb C \backslash \{0,1\}$ with $f(0) = a$ and $f'(0)= b$. But then by Montel's three value theorem (cf. Barry Simon's complex analysis book), the limit function $f$, which must not be infinity because $f_n(0) = a < \infty$, is an entire function that misses $0,1,$, and $\infty$, and such that $f_n$ converges uniformly on every compact subset of $\mathbb C$. But this uniform convergence and the fact that $f, f_n$ are holomorphic imply that on any compact set $f_n'(0) = b \neq 0$ converges to $f'(0) = 0$, which is impossible.
Thus there must exist some $R > 0$ such that there is no holomorphic function satisfying the conditions.
EDIT: the following before my edit of the above answer seems unnecessary and has a logical fallacy, please ignore this: "thus, we have shown that there is some $R \geq 0$ such that the result holds true. It remains to show that $R \neq 0$. The function $g(z):= a + bz$ is a holomorphic function on the ball $\{z: |z| < \min(|-\frac{a}{2b}|, |\frac{1-a}{2b}| \})$ that cannot take values $0$ and $1$ and such that it satisfies the conditions. Any $R > 0$ such that $R < \min(|-\frac{a}{2b}|, |\frac{1-a}{2b}| \})$ has $g|_R$ as a holomorphic function that satisfies the conditions. Thus, $R \neq 0$."
The proof is essentially correct. Of course one must require $b \ne 0$ in the statement. Also Montel's theorem only guarantees locally uniform convergence of a subsequence of the original sequence.
A technical “problem” is that the functions $(f_n)$ do not have a common domain. This can be “solved” as follows:
This gives a double sequence $(f_{k, n})$ with the property that for each $k$, $(f_{k, n})_n$ is defined and locally uniformly convergent in $|z| < k$. The “diagonal sequence” $F_n(z) = f_{n, n}(z)$ then has the property that for each $k$, $(F_n)_{n \ge k}$ is defined and locally uniformly convergent in $|z| < k$, so that we can define the limit function $$ f(z) = \lim_{n \to \infty} F_n(z) $$ for all $z \in \Bbb C$.
The limit function $f$ is not constant because $f'(0) = b \ne 0$, so that it is a non-constant entire function which misses the values $0, 1$, which contradicts Picard's theorem.