Let $f,g,h$ be "nice" functions and $h(H(x))=H(h(x))=x$. Let $F(x)=H(f(h(x)))$ and $G(x)=H(g(h(x)))$. Obviously $f(g(x))=g(f(x)) \Leftrightarrow F(G(x))=G(F(x))$.  

Example: $f(x)=ax, g(x)=bx$ and $F(x)=a+x,G(x)=b+x$ are related by a log transformation.  

Question: Can all commuting functions $f,g$ be transformed to $F(x)=a+x,G(x)=b+x$ with some $h$? You may restrict "nice" as you like (monotonous? differentiable? complex domain?) if a proof demands it. (It's probably not hard to come up with an un-nice counterexample.)



Are all commuting functions "the same"?
Let me rephrase your question: is every pair of commuting functions simultaneously conjugate to
a pair shifts?
(Your conditions, literally understood imply that your $h$ is a bijective map.)
The answer is evidently no.


In your example, $ax$ and $bx$ are restricted to the positive ray.
If you consider the "same" functions on the whole real line, they are already
not conjugate to shifts. Because they have a fixed point while the shifts do not.


One can replace conjugacy by semiconjugacy, requiring in your notation $h(F)=f(h)$,
and $h(G)=g(h)$ (without the requirement that $h$ is invertible) but even in this case, not every pair of commuting functions is
semiconjugate to an affine pair.


Commuting rational functions are almost completely classified in the paper of J. F. Ritt,
TAMS 1923, and another proof of the same result can be found in my paper MR1027462.
Only for polynomials (and some subclasses of rational functions, like Blaschke products)
the problem of description of all commuting pairs is completely solved.


There is an enormous literature about this and other classes of commuting functions
and maps. Under some additonal conditions, like having a common fixed point,
commuting pairs are frequently (not always) semiconjugate to pairs of linear maps. 


