Proving the area of a square and the required axioms
The question is where do you start with the definition of "area". As far as I know, there are two options:
Develop a general notion of a measure in the plane, say, in the spirit of the Lebesgue integral. Then the formula for the measure of a rectangle comes as a part of the definition of the measure. The hard part of the theory is to construct a well-defined set-function on a certain algebra of subset of the plane (or $n$-space, there is no fundamental difference between the two constructions).
You can also start with axioms of Euclidean geometry (which do not include anything about areas!) and then develop a concept of area for general (not only convex) polygonal regions in the plane. If your objective is to present something that high school students or typical undergraduate students (in a Euclidean geometry class) can comprehend, then this is a way to go. Take a look at the book
E. Moise, "Elementary geometry from advanced standpoint", especially section 13.5, where he shows that area of any square (with side $a$) is $a^2$, starting merely with the normalization that area of the unit square is 1. Along the way, Moise develops the notion of area of planar polygons from scratch. The most difficult part of his proof (that area exists) is to show that area of a polygon does not depend on triangulation.
To conclude: all you need is the set of (Hilbert's) axioms of Euclidean geometry (since, as Hilbert observed, Euclid's set of axioms is incomplete). You also need a notion of real numbers (which Euclid did not have either), namely, ordered field axioms and the completeness axiom.
One very practical and natural way of approaching area is to give the following definition:
The area of a plane figure is the number of unit shapes required to cover it up completely with no overlapping or going over the edges.
We usually take the unit shape to be a square of side length $1$. Note that we have to assume that area is well-defined, in other words that we couldn't cover a figure exactly with $9$ squares, and then cover the same figure exactly with $12$ squares in a different arrangement. Intuitively this seems absurd, and we have to accept it as an axiom here. From the definition we can immediately see that the area of two combined but disjoint figures is the sum of the areas of the individual figures.
We can already prove the formula for the area of a square whose sides are of integer length by covering the square in the obvious way, but what about for fractional lengths? Actually, for fractional lengths, the area might well be undefined. Consider a square of side length $1/2$. No matter how hard you try, you won't be able to cover that up exactly with unit squares, so according to the above definition, it doesn't really have an area.
For simplicity, consider a rectangle $R$ of dimensions $1/2\times1$. What is its area? $1/2$? Why? Well, surely it's one-half... after all, $R$ is simply a unit square, cut in half... but wait, what do you mean by "cut in half"? We're going in circles.
The solution is to go to a smaller unit of measurement, extending the definition:
If it takes $n$ copies of a shape to cover a unit square, the shape has area $1/n$. If a shape can be covered exactly by any shapes at all, its area is that sum of the areas of those shapes.
So again, what is the area of a square of side length $a/b$ ? Well, such a square can be broken up into $a^2$ little squares of side length $1/b$, and it's obvious that you can arrange $b^2$ such little squares to cover a unit square exactly, thus by the above definition such a square has side length $1/b^2$. Therefore the area of the original square is $a^2\cdot1/b^2$.