Letting $\eta = e^{1/e}$ where $e$ is Euler's constant, there exists a function $F(z)=\, ^z \eta$ with the following relevant properties. (I won't bother showing the existence of this function, or the construction, but it is there in the cosmos of function space.)

Let $z= x+iy$

$$F:\mathbb{C}_{x>-2} \to \mathbb{C}$$

$$F(0) = 1$$

$$\eta^{F(z)} = F(z+1)$$

$$F:(-2,\infty) \to (-\infty,e)$$

$$\frac{d}{dz}F(z) \neq 0$$

$$\frac{d}{dx}F(x) > 0$$

This is tetration base $\eta$, and is a valuable function for many reasons. We are sticking to the regular iteration method.

Now my question is difficult to state but revolves around extending this function to a larger domain than the half plane $\Re(z) > -2$. It is rather trivial to show $$\log_\eta(F(z)) = e\log(F(z)) = F(z-1) + e2\pi i k$$ for an arbitrary branch of $\log$, for some $k \in \mathbb{Z}$--locally about a point $z$.

What I'm trying to understand is if the following is an extension of $F$ for $|\arg(z-2)| < \pi$. Namely if $F$ extends everywhere to the complex plane excepting $y=0,x\le-2$. It is difficult to state but the proof of an extension goes as follows.

$F(-1) =0$ and therefore $\log(F(z))$ is not holomorphic in a neighborhood of $z=-1$ so that $F(z)$ is not holomorphic for $z$ in a neighborhood of $-2$ necessarily. So that the singularity at $-2$ is a logarithmic singularity--a branch point if you will.

Using the functional equation of $F$: $\eta^{F(z)} = F(z+1)$ and by taking the implicit equation

$$\eta^{\mu} - F(z-2)=0$$

for $0 < x < 1+\epsilon,\,y>0$ and arbitrary $\epsilon > 0$, we have a bunch of functions $\mu$ defined locally about each point $z = x+ i y$. Now $\mu + e2\pi i k $ is equally so a plausible function by the periodicity of $\eta^z$.

What grounds do we need so that we can say there is $\mu$ holomorphic for all $0 < x < 1 + \epsilon,\,\,y>0$? As in, what do we need to show in order to say this covering of the domain with analytic functions $\mu$ about each point $z$ forms an analytic function on the entire domain?

I assume the result requires some knowledge in Riemann surfaces (something I am a beginner in), and I assume it is not trivial to show this result. Multivalued functions have never been my strong suit, and so I hope the answer can be found in simpler terms than some advanced argument from complex manifolds.

If $\mu$ is holomorphic in this domain, there is only one $\mu$ that agrees with $F(z-3)$ for $1 < x < 1 + \epsilon$ by the identity theorem. And therefore we have found an extension of $F(z)$. Repeating this procedure eventually shows $F(z)$ is holomorphic for all $y>0$, and an equal procedure works for $y<0$. This in the end shows $F(z)$ is holomorphic everywhere excluding $y=0,x\le-2$--i.e: when $|\arg(z-2)|<\pi$.