I think I can prove that $ab = ba$ by other axioms. Am I wrong? Why?
Edit: proof updated with notes showing where I actually used the commutative property without noticing.
Proof. For any numbers $a$ and $b$, we know this is true: $$ ab + ab = ab + ab $$
By using the distribution axiom: $$\begin{split} a(b + b) &= \underbrace{b(a + a)}_{\text{mistake: implies commutivity}}\\ a(2b) &= b(2a)\\ a2b &= b2a\\ \end{split}$$
By multiplying both sides by $2^{-1}$: $$\begin{split} a2b2^{-1} &= b2a2^{-1}\\ \end{split}$$
By the multiplication associative axiom: $$\begin{split} ab(22^{-1}) &= ba(22^{-1}) \qquad\text{(mistake: requires commutivity)}\\ \end{split}$$
By the multiplication inverse axiom: $$\begin{split} ab(1) &= ba(1)\\ \end{split}$$
By the multiplication identity axiom: $$ab = ba$$
$\blacksquare$
Two things. If you don't have commutativity we can't claim $b(a+a) = (a+a)b,$ and so distribution would allow only $ab +ab = (a+a)b= 2ab$ and you can't get $ab+ab =b(a+a)= b2a$.
Second you claim you can get $a2b2^{-1}= ab(22^{-1})$ from associativity. I... just can't see any justification as to how associativity implies that, as the $b$ and the $2$ have "switched places"... I can't see any reason that could be without commutivity.
....
Interestingly though for all the "integers" -- terms that arise as $1+1 = 2$ and $2+1 = 3$, etc. -- you can claim limited commutativity as
$2b = (1+1)b = b+b = b(1+1) = b2$.
But I don't think you can extend that to elements that are not "rational". (i.e., not a sum of $1$s or a sum of $1$ multiplied by an inverse of another sum of squares). But I think you can prove by induction all "rationals" in a "non-commutative field" commute.
But I don't think a non-commutative field, in and of itself, is internally inconsistent. (I could be wrong.)