Can we define addition, subtraction, and multiplication operators as a relation using sets on Natural Numbers? Or are they considered to be axioms, and don't have any definition in terms of a set?
 A: Yes, this is possible. Here is an outline of how one might do that:
We start by defining the set of natural numbers as the $\subseteq$-minimal transitive set $N$ such that


*

*$\emptyset \in N$ and

*$N$ is closed under $n \mapsto n \cup \{n \} =: S(n)$.


Now $(N; \in)$ is a strictly ordered countable set and $N$ should be thought of as $\mathbb N$ (or a copy thereof). Intuitively $^{(\dagger)}$ $N$ looks like this:
$$
N = \{\underbrace{\emptyset}_{=0}, \underbrace{S(\emptyset)}_{=1}, \ldots, \underbrace{S^n(\emptyset)}_{=n}, \ldots \}.
$$
We recursively define addition on $N$ by


*

*$n + \emptyset = n + 0 = n$ and

*$n + S(m) = n + (m+1) = (n+m) + 1 = S(n+m)$.


By the recursion theorem this yields a well-defined function
$$
+ \colon N \times N \to N, (n,m) \mapsto n + m.
$$
In a similar fashion one can define subtraction (defining $n-m := 0$ whenever $n < m$) and multiplication on $N$.
One can then use the structure $(N; <, +, \cdot, -, 0,1)$ to define the set of integers as some quotient of pairs of natural numbers. And one can use the addition and multiplication on the naturals to equip the integers with their natural structure.
Pushing forward, one can now use the integers to define the rationals, again as a quotient of pairs of integers. And once again one can use the structure on the integers to equip the rationals with their natural structure.
Using rational, we then can define reals - as quotients of countable sequences of rationals. And, with only slightly more trouble then before, one can naturally lift the structure of the rationals to equip the reals with their natural structure.
Once we have the reals, the usual construction allows us to define complex numbers, all sorts of vector fields, ...
All of this can be done in quite weak set theories and doesn't require the axiom of choice at all. (Proving certain statements about these structures, however, sometimes does need significantly stronger theories and does explicitly require (or forbid) some level of choice.)
$(\dagger)$ I don't think this is the right place to get into this. But, more or less as a consequence of the compactness theorem, this intuition isn't entirely correct.
