Some relevant background (all this can be found in a survey by Aschenbrenner and van den Dries on aymptotic differential algebra): the closure of $\exp, \log$ and real constants under field operations and composition is called the class of logarithmic-exponential functions, or LE-functions for short. It is a fundamental example of a Hardy field, which by definition is a subalgebra of the algebra of germs at infinity of $C^\infty$ functions that (1) forms a field, and (2) is closed under the derivative operation $f \mapsto f'$.
(Notes: (1) an LE-function is uniquely determined by its germ at infinity by analytic continuation. (2) In order for such a subalgebra $H$ of germs $[f]$ of functions $f$ to form a field, we need that each $f$ has eventually constant sign in $\{-,0, +\}$, and the same for all its derivatives by closure under differentiation. Hardy fields are thus ordered differential fields containing the field $\mathbb{R}$.)
We say that a (germ) $f$ is infinite if $|f|$ is not bounded above by any constant $r \in \mathbb{R}$.
Lemma: If $f$ defines an infinite germ in a Hardy field and $f'$ is (eventually) positive, then its compositional inverse $f^{-1}$ also belongs to a Hardy field (a larger one where $f^{-1}$ 1 is also infinite of course).
This is Theorem 1.7 in the linked reference by Aschenbrenner and van den Dries.
The question of the OP on whether every $f \in E$ is asymptotic to an LE-function is settled negatively by exhibiting a suitable LE-function $f$ whose compositional inverse is not asymptotic to an LE-function (suitable meaning that the growth rate of the inverse is not too extreme, as required in the OP). Hardy himself conjectured that $f(x) = (\log x)(\log \log x)$ should work. This was finally shown here:
- L. van den Dries, A. Macintyre, D. Marker, Logarithmic-exponential power series, J. London Math. Soc. 56 (1997), 417–434.
Another example of an $f$ that works is $f(x) = (\log \log x)(\log \log \log x)$, as shown by Shackell in 1993; see reference [68] in the survey. I've just now learned that Shackell has written a whole book in this area, titled Symbolic Asymptotics.
I've only done a little amateur dabbling in this area, and so I am absolutely in no position to give a full answer to the OP's question, but $E$ is quite enormous, and for what it's worth I doubt that any easily described set of functions generates (under algebraic operations and composition) the full set of asymptotic equivalency classes for $E$. See the cited survey for some useful information on various huge Hardy fields. One area of current interest is the nature of the intersection of all maximal Hardy fields (there's a Zorn's lemma argument that every Hardy field is contained in a maximal Hardy field); google the name Michael Boshernitzan for more on this.
As for the bonus question: for functions $f, g$ belonging to a Hardy field $H$, it's pretty much a triviality. Put $h = f/g$, and let $r = \limsup h(x) \in [0, \infty]$. That is, put $r = \inf_x \sup_{y \geq x} h(y)$. Then show $h(x)$ converges to $r$ by considering cases ($h$ eventually increasing, $h$ eventually decreasing, $h$ eventually constant). The trouble with the class $E$ of the OP is that it's not (I don't think) a Hardy field and hence isn't under very good algebraic control. For an example to consider, I'd think that something like $f(x) = \exp(\exp (x) + \sin (x))$ has all the derivative requirements of the OP [the $n^{th}$ derivative has a term $f(x)(e^x + \cos(x))^n$ which eventually dominates the other terms], and so does $g(x) = \exp(\exp (x))$, but the quotient $f/g$ has oscillatory behavior.
Edit: As pointed out by Gérard H.E. Duchamp in a comment below, Bourbaki also gave a nice treatment of Hardy fields (Functions of a Real Variable, published by Springer). In particular, the issue of compositional inverses is given in an appendix (paragraph 6).