$A$ be an elementary function, algebraic over $\mathbb{C}$,
$f_1$ and $f_2$ are bijective elementary functions with elementary inverses,
$F\colon z\mapsto A(f_1(f_2(z)))$ be a bijective elementary function.
What is the inverse $F^{-1}$ of $F$?
See Wikipedia: Elementary function, consider the definition of the elementary functions of Liouville and Ritt.
Ritt wrote in Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90: "That every $F(z)$ of this type has an elementary inverse is obvious."
Is this really obvious? If all mentioned functions are bijective, $F^{-1}=f_2^{-1}\circ f_1^{-1}\circ A^{-1}$. But only injectivity of $f_2$ and surjectivity of $A$ follow from the bijectivity of $F$. Therefore $A$ may be non-injective.
Bijectivity of $A$ can be defined by restriction and corestriction. But are this new function and its inverse (a partial inverse of the original $A$) still elementary functions? They may have different domain and/or codomain in comparison to the original elementary functions. It depends on the answer to the question "Are all restrictions of an elementary function also elementary functions?".
And what if $A$ is not bijective: can its bijective restriction to the codomain of $f_1\circ f_2$ represented by a $single$ bijective elementary function?
If $f_1$ and $f_2$ and $A\circ f_1 \circ f_2$ are all bijective $X\to X$ then necessarily $A$ will be bijective too.
If $A$ fails to be either injective or surjective, this failure would be inherited by $A\circ f_1 \circ f_2$. This is obvious for surjectivity; for injectivity it is a consequence of $f_2$ and $f_2$ both being surjective.