Given a function $f(x)$ with a taylor series expansion, is it valid to say that $$f(x)=\sum_{n=0}^{\infty}\frac{1}{n!}f^{(n)}(a)(x-a)^n$$ for all values of x irrespective of whether the taylor series converges for all x? In other words, if we were to sum up all the infinite terms of the taylor series all at once, should we end up with f(x), irrespective of whether the series converges for that particular value of x?
But the action of summing an infinite number of things all at once is similar to the Axiom of Choice in the respect that it is doing an infinite action of taking something from each of infinite number of things.
So the question boils down to the question of whether there already exists, or whether there could exist an axiom which allows us to do that action of infinite summation which I have in mind? Would it be related to the Axiom of Choice? And if it exists should we really end up with the equivalence between the function and its taylor expansion?
Infinite summation has a specific definition which does not use the axiom of choice. Just the definition of a sequential limit. It is often beneficial to think about an infinite sum as actually a summation of infinitely many numbers, but in reality it is really just a limit of finite summations.
You don't need the axiom of choice for this, since the sequence of partial sums is well-defined and requires no arbitrary choices; and the definition of a convergent sequence has nothing to do with the axiom of choice.
You also don't need the axiom of choice to show that an analytic function has a Taylor expansion, since the expansion is defined from the function in a very specific way.