Lets summarize what we have so far:
Proposition. Let \( S \) be a vertical strip somewhat wider than \( 1 \), i.e. \( S=\{z\in\mathbb{C}: x_1-\epsilon<\Re(z)<x_1+1\} \) for some \( x_1\in\mathbb{R} \) and \( \epsilon>0 \).
Let \( D=\mathbb{C}\setminus (-\infty,x_0] \) for some \( x_0<x_1 \) and let \( G\subseteq G' \) be two domains (open and connected) for values, and let \( F \) be holomorphic on \( G' \). Then there is at most one function \( f \) that satisifies
(1) \( f \) is holomorphic on \( D \) and \( f(S)\subseteq G\subseteq f(D)=G' \)
(2) \( f \) is real and strictly increasing on \( \mathbb{R}\cap S \)
(3) \( f(z+1)=F(f(z)) \) for all \( z\in D \) and \( f(x_1)=y_1 \)
(4) There exists an inverse holomorphic function \( f^{-1} \) on \( G \), i.e. a holomorphic function such that \( f(f^{-1}(z))=z=f^{-1}(f(z)) \) for all \( z\in G \).
Proof. Let \( g,h \) be two function that satisfy the above conditions. Then the function \( \delta(z)=g^{-1}(h(z)) \) is holomorphic on \( S \) (because \( h(S)\subseteq G \) and (4)) and satisfies \( h(z)=g(\delta(z)) \). By (3) and (4)
\( \delta(z+1)=g^{-1}(F(h(z)))=g^{-1}(F(g(\delta(z))))=g^{-1}(g(\delta(z)+1))=\delta(z)+1 \) and \( \delta(0)=0 \).
So \( \delta \) can be continued from \( S \) to an entire function and is real and strictly increasing on the real axis.
\( \delta(\mathbb{C})=\mathbb{C} \) by our previous considerations.
By Big Picard every real value of \( \delta \) is taken on infinitely often if \( \delta \) is not a polynomial, but every real value is only taken on once on the real axis, thatswhy still \( \delta(\mathbb{C}\setminus\mathbb{R})=\mathbb{C} \). But this is in contradiction to \( g^{-1}:G\to D=\mathbb{C}\setminus [x_0,\infty) \). So \( \delta \) must be a polynomial that takes on every real value at most once. This is only possible for \( \delta(x)=x+c \) with \( c=0 \) because \( \delta(0)=0 \).\( \boxdot \).
In the case of tetration one surely would chose \( x_0=-2 \) and \( x_1=0 \) or \( x_1=-1 \). However I am not sure about the domain \( G \) which must contain \( f(S) \) and hence give some bijection \( f:G\leftrightarrow S' \), with some \( S'\supseteq S \).
Of course in the simplest case one just chooses \( G=f(S) \) if one has some function \( f \) in mind already. However then we can have a different function \( f_2 \) with \( f(S)\neq f_2(S) \) but our intention was to have a criterion that singles out other solutions.
So we need an area \( G \) on which every slog should be defined at least and satisfy \( \text{sexp}(\text{slog})=\text{id}=\text{slog}(\text{sexp}) \) as well as \( \text{sexp}(S)\subseteq G \).
Proposition. Let \( S \) be a vertical strip somewhat wider than \( 1 \), i.e. \( S=\{z\in\mathbb{C}: x_1-\epsilon<\Re(z)<x_1+1\} \) for some \( x_1\in\mathbb{R} \) and \( \epsilon>0 \).
Let \( D=\mathbb{C}\setminus (-\infty,x_0] \) for some \( x_0<x_1 \) and let \( G\subseteq G' \) be two domains (open and connected) for values, and let \( F \) be holomorphic on \( G' \). Then there is at most one function \( f \) that satisifies
(1) \( f \) is holomorphic on \( D \) and \( f(S)\subseteq G\subseteq f(D)=G' \)
(2) \( f \) is real and strictly increasing on \( \mathbb{R}\cap S \)
(3) \( f(z+1)=F(f(z)) \) for all \( z\in D \) and \( f(x_1)=y_1 \)
(4) There exists an inverse holomorphic function \( f^{-1} \) on \( G \), i.e. a holomorphic function such that \( f(f^{-1}(z))=z=f^{-1}(f(z)) \) for all \( z\in G \).
Proof. Let \( g,h \) be two function that satisfy the above conditions. Then the function \( \delta(z)=g^{-1}(h(z)) \) is holomorphic on \( S \) (because \( h(S)\subseteq G \) and (4)) and satisfies \( h(z)=g(\delta(z)) \). By (3) and (4)
\( \delta(z+1)=g^{-1}(F(h(z)))=g^{-1}(F(g(\delta(z))))=g^{-1}(g(\delta(z)+1))=\delta(z)+1 \) and \( \delta(0)=0 \).
So \( \delta \) can be continued from \( S \) to an entire function and is real and strictly increasing on the real axis.
\( \delta(\mathbb{C})=\mathbb{C} \) by our previous considerations.
By Big Picard every real value of \( \delta \) is taken on infinitely often if \( \delta \) is not a polynomial, but every real value is only taken on once on the real axis, thatswhy still \( \delta(\mathbb{C}\setminus\mathbb{R})=\mathbb{C} \). But this is in contradiction to \( g^{-1}:G\to D=\mathbb{C}\setminus [x_0,\infty) \). So \( \delta \) must be a polynomial that takes on every real value at most once. This is only possible for \( \delta(x)=x+c \) with \( c=0 \) because \( \delta(0)=0 \).\( \boxdot \).
In the case of tetration one surely would chose \( x_0=-2 \) and \( x_1=0 \) or \( x_1=-1 \). However I am not sure about the domain \( G \) which must contain \( f(S) \) and hence give some bijection \( f:G\leftrightarrow S' \), with some \( S'\supseteq S \).
Of course in the simplest case one just chooses \( G=f(S) \) if one has some function \( f \) in mind already. However then we can have a different function \( f_2 \) with \( f(S)\neq f_2(S) \) but our intention was to have a criterion that singles out other solutions.
So we need an area \( G \) on which every slog should be defined at least and satisfy \( \text{sexp}(\text{slog})=\text{id}=\text{slog}(\text{sexp}) \) as well as \( \text{sexp}(S)\subseteq G \).