On extension to "other" iteration roots

[Leo.W]

but we in everyday talk and in forum it's default to say something taking singlevalued-ness for granted,

On the other hand, in the complex plane, it is rather the default to consider multivaluedness. E.g. log or sqrt, when you continue from one point around the singularity back to the same point, you arrive at a different value and that's also the secret with the iteration group, it is not so much having multiple separate values, they just come naturally into existence by analytic continuation (and that's why the graph is one line not many separate lines).

Apropos Abelian property: just want to remind you to be cautious with the term, because \(f^{s}\circ f^{t} = f^{t}\circ f^{s}\) does not automatically imply \(f^{s}\circ f^{t} = f^{s+t}\) if the latter is what you actually mean.
I say that because I just encountered that case with the real valued Fibonacci extension
\begin{align}
\phi'_t &:=\frac{\Phi^t+\cos(\pi t)(-\Psi)^t}{\Phi-\Psi}\\
f^{t}(z) &:= \frac{\phi'_t + \phi'_{t-1}z}{\phi'_{t+1} + \phi'_t z}\\
\end{align}
There we have \(f^{s}\circ f^{t} = f^{t}\circ f^{s}\) but we don't have \(f^{s}\circ f^{t} = f^{s+t}\) for most s,t and hence it is not a continous iteration group.
And that would actually be your counterexample to have a real single valued function family, with the Abelian property at a fixed point with negative multiplier

Not to disappoint ya, bo, but you'd missed one thing is that, you proved the \(f^{s}\circ f^{t} = f^{t}\circ f^{s}\) identity taken its singlevaluedness, this will be the same as:

Q: Find an abelian realvalued (singlevalued) superfunction for \(f(z)=z+1\)
A: It's okay you write \(f^t(z)=z+g(t)\) where g is any function with \(g(n)=n\) for integers because \(g(s)+g(t)=g(t)+g(s)\)
Thus there lies such a superfunction \(F(z+1)=f(F(z))=F(z)+g(t)\), \(F(z)=z+G(t)\) is a superfunction where \(G(t)=F(0)+\sum_{n=0}^{t-1}{g(n)}\), choose a real-valued arbitrary function g(n) with \(g(n)=n\) for integers n, will always promise you such a superfunction and abelian-ity.

But the core is, Abelian-ity for iterations must refers to \(f^{s+t}=f^s\circ f^t=f^t\circ f^s\), otherwise we're just doing almost the same thing as finding "All matrices A that commutes with B", it not suffices to call itself an iteration, just an Abelian group of functions that commutes with each other.

If you look into the multivaluedness of your superfunction, then your superfunction can preserve an Abelian-ity by some branch cuts.
Assume your superfunction \(F\) has a multivalued inverse \(H\), then since \(H(f)=H+1, F(z+1)=f(F)\), by conjugacy \(H(F(z)+1)=f(z)\), thus a naturally extended iterations \(H(F(z)+t)=f^t(z)\) and satisfy Abelianity. All equalities here represents "There lies such a branch which fits this equality".

Another thing, your extended fibonacci is used for so long, it's derived from a complex decomposition by Re and Im:
Assume \(f(z+1)=L(f(z),f(z-1),f(z-2),\cdots,f(z-n))\) has a complexvalued solution f where L is a linear function, then
\(\Re(f(z+1))=\Re(L(f(z),f(z-1),f(z-2),\cdots,f(z-n)))=L(\Re(f(z)),\Re(f(z-1)),\cdots,\Re(f(z-n)))\)
also true for replacing \(\Re\) with \(\Im\), So \(\Re(f(z))\) and \(\Im(f(z))\) are also solutions, your superfunction can be derived from \(\Re\) case.
It may be possible for you to get a real singlevalued/multivalued superfunction, try on this:
(since the hidden eigen equation is \(f(z+1)=f(z)+f(z-1)\) or fibonacci, it must follows any \(c_tf(z+t)\) satisfies the equation. We even can write \(G(z)=\int_{a}^{b}{f(z+t)a(t)\mathrm{d}t}\) or \(H(z)=\sum_{n=a}^{b-1}(c_nf(z+n))\) as extended solution, where a(t) arbitrary function, a,b arbitrary constant. It's almost the same as writing a series, or almost equivalently a thetamapping. But intrinsically it's only) assuming \(G(z)=a_1(t)r_1^z+a_2(t)r_2^z\). where \(r_1,r_2\) solves \(r^2-r-1=0\)