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 $|\arg(z-2)|<\pi$.



Branches of the tetration function
You are looking at the solution of the **Abel functional equation** for your function
$$ f(z) = \eta^z = e^{\frac{z}{e}}.$$
(Note that this function is conformally conjugate to the map $\zeta \mapsto e^\zeta-1$ via $z = e\cdot (\zeta +1)$, which may be more convenient as its fixed point is at $0$ instead of $e$.)


This function has a fixed point at $z=e$ with derivative $1$. The solution of the Abel functional equation for such functions is well-understood, and referred to as the **Fatou coordinate**. In particular, you can extend the map by analytic extension along horizontal lines to all of
$$\newcommand{\C}{\mathbb{C}}\C\setminus (-\infty,-2]$$
(where I am using your normalisation, where the function maps $-1$ to the omitted value at $0$, although I am not sure why you would use this normalisation). 


All you need to know to justify this is that the map $f$ is a covering map $f\colon \C\to \C\setminus \{0\}$ (so that the lift / analytic continuation is defined), and that your domain is simply-connected so that you can use the monodromy theorem.


As you note, you cannot extend the function continuously to any point of $(-\infty,-2]$. However, the \emph{inverse} $\phi$ of your function $F$ can easily be extended, via the functional relation, to the entire **basin of attraction**, i.e. all points converging to the fixed point $e$ under iteration of $f$. 


This gives an alternative view on your result: The interval $(-\infty,-2]$ contains all the **singular values** of this function $\phi$ (i.e., values that don't have a neighbourhood over which $\phi$ is a covering map). So what you are doing is taking a branch cut through all the singular values of $\phi$, and taking one of the (infinitely many) branches of $\phi^{-1}$ defined on this set.


