The simplest and historical answer is that we observed that certain things in the world can be counted, in the sense that we can put them in a line and label each one with their position in the line, and that these positions satisfy certain empirical facts, and hence we invented axioms to capture these facts. I specifically single out the axioms in that linked section (discrete ordered semi-ring axioms plus induction), rather than the other more commonly known axiomatization of PA, for two reasons.
Firstly, the ring axioms have been observed or proven to hold for a very wide variety of structures, so it is no surprise that people invent them, and they include distributivity of multiplication over addition.
Secondly, that specific axiomatization I linked to accurately captures our empirical knowledge of counting numbers. For example, from our understanding of spatial motions (translations and rotations) we readily grasp that addition is commutative: $a+b$ is the total count of $a$ objects followed by $b$ objects in one line, which is the same as $b+a$ by looking at the line from the other side. Similarly for the commutativity of multiplication: $a·b$ is the total count of a rectangular array of $a$ times $b$ objects, which is equal to $b·a$ by rotating the array by $90$ degrees. And distributivity ($a·(b+c) = a·b+a·c$) is clear by splitting the array. Associativity is also intuitive.
Note that the above axioms are sufficient for us to recover general distributivity of the form you are using: Observe that $(a+b)·c = c·(a+b) = c·a+c·b = a·c+b·c$ by commutativity and distributivity. Thus $(a+b)·(c+d) = (a+b)·c + (a+b)·d = (a·c+b·c) + (a·d+b·d)$, and the order of addition in the last expression does not matter by associativity.
There is a more complicated answer, that also goes a significant way to explaining why counting numbers are the way they are. If you have an operator $f$ on some collection $S$ that can be iterated (i.e. $f : S \to S$), then you can define $f^0 = \text{id}_S$ and $f^1 = f$, and define $f^{m+n} = f^m \circ f^n$ and $f^{m·n} = (f^m)^n$ for every counting numbers $m,n$. The last definition is valid because (by induction) $f^m$ is an operator on $S$ that can be iterated. Then (again by induction), you can actually prove the basic properties of $+,·$. For example:
Associativity of $+$
Take any naturals $k,m,n$. Then $f^{k+(m+n)} = f^k \circ ( f^m \circ f^n ) = ( f^k \circ f^m ) \circ f^n = f^{(k+m)+n}$. Here we are using the fact that function composition is associative, which is a core reason for almost every instance of associativity in mathematics.
Commutativity of $+$
First we show that $f^{m+1} = f^{1+m}$ for every natural $m$. $f^{0+1} = f^0 \circ f = f = f$ $\circ f^0 = f^{1+0}$ by definition. For any natural $n$ such that $f^{n+1} = f^{1+n}$, we also have $f^{(n+1)+1} = f^{n+1} \circ f = f^{1+n} \circ f$ $= f^{(1+n)+1} = f^{1+(n+1)}$. Therefore by induction we are done.
Now take any natural $m$. Then $f^{m+0} = f^m \circ f^0 = f^m$ $= f^0 \circ f^m = f^{0+m}$. And for any natural $n$ such that $f^{m+n} = f^{n+m}$, we also have $f^{m+(n+1)} = f^{(m+n)+1} = f^{m+n} \circ f = f^{n+m} \circ f$ $= f^{(n+m)+1} = f^{n+(m+1)}$ $= f^n \circ f^{m+1} = f^n \circ f^{1+m}$ $= f^{n+(1+m)} = f^{(n+1)+m}$. Therefore by induction we are done.
As you can see from the above proofs, induction is enough for us to bootstrap from very rudimentary notions of iteration to obtain addition and its properties. We can do the same for multiplication. But it should be clear that this was not how the axioms were originally invented.