On the commutative property of multiplication (domain of integers, possibly reals)

The Peano axioms provide an axiomatic basis for the natural numbers, including addition and multiplication of them. In short, they define $0$ and a successor function $S$ which is used to define the natural numbers by $1:=S(0),2:=S(1),\ldots$ in the obvious manner. Addition is defined recursively by $a+0=a$ and $a+S(b)=S(a+b)$, while multiplication is defined by $a\cdot 0=0$ and $a\cdot S(b)=a+(a\cdot b)$. From these axioms and induction we can prove that addition and multiplication are commutative. For my proof I will assume that addition is both associative and commutative, but if you want you can prove these facts from the axioms as well.

Lemma 1: $0\cdot b = 0$ for all $b$.

Proof: For the base case, $0\cdot 0=0$ by the axioms. If $0\cdot b=0$, then $0\cdot S(b)=0+(0\cdot b)=0+0=0$. Thus by induction $0\cdot b=0$ for all $b$.

Lemma 2: $a+(b\cdot a)=S(b)\cdot a$ for all $a,b$.

Proof: We shall induct on $a$. For the base case, $0+(b\cdot 0)=b\cdot 0=0=S(b)\cdot 0$. Suppose $a+(b\cdot a)=S(b)\cdot a$. Then $$\begin{eqnarray} S(a)+(b\cdot S(a))&=&S(a)+(b+(b\cdot a))\\ &=&(S(a)+b)+(b\cdot a)\\ &=&(b+S(a))+(b\cdot a)\\ &=&S(b+a)+(b\cdot a)\\ &=&S(a+b)+(b\cdot a)\\ &=&(a+S(b))+(b\cdot a)\\ &=&(S(b)+a)+(b\cdot a)\\ &=&S(b)+(a+(b\cdot a))\\ &=&S(b)+(S(b)\cdot a)\\ &=&S(b)\cdot S(a) \end{eqnarray}$$ so by induction this holds for all $a,b$.

Proof of Theorem: For the base case, $a\cdot 0 = 0 = 0\cdot a$. If $a\cdot b=b\cdot a$, then $a\cdot S(b)=a+(a\cdot b)=a+(b\cdot a)=S(b)\cdot a$. Thus by induction $a\cdot b=b\cdot a $ for all $a,b$.