Partial answer.
- If $f$ is a composition, then the last composed functions can be ignored if they are injective. Thus, for anything of the form $e^{h},h^{2n+1},\ln h,\ldots$, we need just consider $h(x)$.
Proof: Suppose we were to construct a periodic, composite function $f(x)=g\circ h(x)$, where $g$ is injective and thus aperiodic. Then $h$ must have already been periodic since it is trivially $h=g^{-1}\circ f$.
- $f$ cannot be a rational function with an injective function as its variable.
Proof: If $f(x)$ were a periodic polynomial then, for some $x_0$, $f(x)-f(x_0)$ would be a polynomial with infinite roots, $x_0+nP$, where $P$ is a period and $n\in\mathbb{Z}$. This would violate the fundamental theorem of algebra. This can be generalised to any rational function $r(x)=\frac{a(x)}{b(x)}$ since the polynomial $a(x)-r(x_0)b(x)$ would have infinite roots, $x_0+nP$. We can then generalise it further to $r\circ h(x)$ for any rational function, $r$, with an injective function, $h$, as its variable since it would have at most as many roots as $r(x)$. So all functions of the form $\frac{e^{2x}-3e^x}{e^{2x}+1},\frac{\ln(x)^2+1}{\ln(x)},\ldots$ cannot be periodic.
The question remains for combinations of different elementary functions such as $\frac{e^x+\ln(x)+\ldots}{x+\ldots}$. A difficulty in the problem is that the composition of two aperiodic, elementary functions can form a periodic function. For example, take $g(x)=\cos\sqrt{x}$ and $h(x)=x^2$. So too can the sum or product of two aperiodic, elementary functions. For instance, take $\left(\sin x+e^x\right)+\left(-e^x\right)$ or $\frac{\sin(x)}{e^x}\cdot e^x$.