Specifically, my research question is to find operator $A: (\mathbb{R}^+\rightarrow\mathbb{R}^+)\rightarrow\mathbb{S}$, where $\mathbb{S}$ is some totally ordered set, such that for $f, g: \mathbb{R}^+\rightarrow\mathbb{R}^+$ A(f) > A(g) iff $\lim_{x \to \infty} \frac{f(x)}{g(x)} = \infty$ and A(f) = A(g) iff $\lim_{x \to \infty} \frac{f(x)}{g(x)}$ is a positive real less than infinity.
I'm wondering if anyone has knows sources regarding this question, or any guidance in general. I found a few properties of the operators so far, but have zero sources since I can't find anything similar.
Edit: in the paper so far I've been using Bachmann–Landau notation, but I used the limit definition since I think it is more clear.
As a general comment, there shouldn't be a very precise description of the type of asymptotic growth that can occur for several reasons:
-the set of real valued functions is large and the variety of asymptotic growth is difficult to fathom, in particular there are theorems by du Bois-Reymond and Hardy that show that one always finds "growth orders" above a countable set of growth order, and between two countable sets as well.
-properties of comparability of real valued (or integer valued) functions like the maximal size of possible chains or the minimal size of cofinal chains may depend on axioms which are independant of ZFC
-there seems to be a sort of undiscernability of functions that asymptotically lie beyond a ray of fast growing functions called hyperexponential functions of finite strength
As stated, there are no such elements. It would imply that for any two such functions $f,g$, we have either $\lim_{+\infty} \frac{f}{g} \in \mathbb{R}$ or $\lim_{+\infty} \frac{g}{f}=+\infty$ or $\lim_{+\infty} \frac{f}{g}=+\infty$. Whereas in reality the limit may not exist.
If you restrict to a set $\mathbb{A}$ of functions for which this trichotomy holds, then you can just take $\mathbb{S}$ to be the quotient of $\mathbb{A}$ set by the relation $f \sim g$ iff $\lim_{+\infty} \frac{f}{g} \in \mathbb{R}$, and define the order $<$ as $[f]<[g]$ iff $\lim_{+\infty} \frac{g}{f}=+\infty$ (where $[\cdot]$ denotes the equivalence class). Of course this doesn't really tell you much about the nature of those orders of growth.
There are things in the litterature that look like that. In particular, Hardy fields are examples of such sets $\mathbb{S}$ where each function $f \in \mathbb{A}$ must be $C^k$ on some interval $[a_k,+\infty[$ depending on $f$ and $k$, and $\mathbb{A}$ must contain $g$ with $[g]=[f']$ for each $f \in \mathbb{A}$. And $\mathbb{S}$ must be stable under pointwise field operations (after taking equivalence classes). So this is what you said plus ordered field operations plus derivation for $\mathbb{S}$.
Certain Hardy fields can be represented as fiels of formal series called transseries, and they can all be seen as fields of surreal numbers, but not in a canonical way. So in a sense in this context the field of surreal numbers works as a sort of universal domain of quantities that quantify asymptotic growth.
The problem of undiscernability is still witnessed the reliance on arbitrary choice (use of AC) to define the embeddings in general, but it still yields a way in which one can talk about certain Hardy fields in a formal way. For instance there is a conjecture (whose proof should appear not too late in the future) that any property of a "maximal Hardy field" (Hardy fields that cannot be extended as larger Hardy fields) stated in the first order language using field operations, the property "being a constant function", the exponential function and the derivation, is true if and only if it is true in the field of so-called logarithmic-exponential transseries. This is a model-theoretic conjecture by Aschenbrenner, van den Dries, and van der Hoeven.
You can find references on Hardy field and this type of question in the works of Hardyand Boschernitzan for instance. I suggest you look at the article On Numbers, Germs, and Transseries which will then direct you to many relevant references on this deep question. It should also help make sense of this messy answer.