= Solution
Fix $z_0\ne0,1$ and take a small disk about it that avoids zero and one. There is a holomorphic logarithm $L$ on this disk. The function $L(z)/(2\pi i)$ cannot be an integer there: if it were an integer, exponentiation would give $z=1$. Thus $f(L(z)/(2\pi i))$ is holomorphic on the disk.
Where the prescribed principal logarithm and $L$ are both evaluated, they differ by $2\pi i k$ for an integer $k$. By the period-one identity, their compositions with $f$ coincide. This remains true on either side of the principal branch cut, and at a point on that cut one may instead choose the logarithm with argument between zero and $2\pi$. The local holomorphic expressions therefore agree with the stated pointwise definition and with each other. This is <periodic holomorphic descent through the exponential map>, proving
$$
\boxed{g\text{ is analytic on }\mathbb C\setminus\{0,1\}.}
$$
The principal logarithm itself is not analytic across its cut; periodicity of the outer function removes that obstruction.
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