The fraction ring (localization) $\rm\,S^{-1} R\,$ is, conceptually, the universal way of adjoining inverses of $\rm\,S\,$ to $\rm\,R.\,$ The simplest way to construct it is $\rm\,S^{-1} R = R[x_i]/(s_i x_i - 1).\,$ This allows one to exploit the universal properties of quotient rings and polynomial rings to quickly construct and derive the basic properties of localizations (avoiding the many tedious verifications always "left for the reader" in the more commonly presented pair approach). For details of this folklore see e.g. the exposition in section 11.1 of Rotman's Advanced Modern Algebra, or Voloch, Rings of fractions the hard way.
Likely Voloch's title is a joke - since said presentation-based method is by far the easiest approach. In fact both Rotman's and Voloch's expositions can be simplified. Namely, the only nonobvious step in this approach is computing the kernel of $\rm\, R\to S^{-1} R,\,$ for which there is a nice trick:
$\quad \begin{eqnarray}\rm n = deg\, f\quad and\quad r &=&\rm (1\!-\!sx)\,f(x) &\Rightarrow&\ \rm f(0) = r\qquad\,\ \ \ via\ \ coef\ x^0 \\
\rm\Rightarrow\ (1\!+\!sx\!+\dots+\!(sx)^n)\, r &=&\rm (1\!-\!(sx)^{n+1})\, f(x) &\Rightarrow&\ \rm f(0)\,s^{n+1}\! = 0\quad via\ \ coef\ x^{n+1} \\
& & &\Rightarrow&\ \rm\quad r\ s^{n+1} = 0
\end{eqnarray}$
Therefore, if $\rm\,s\,$ is not a zero-divisor, then $\rm\,r = 0,\,$ so $\rm\, R\to S^{-1} R\,$ is an injection.
For cultural background, for an outstanding introduction to universal ideas see George Bergman's An Invitation to General Algebra and Universal Constructions.
You might also find illuminating Paul Cohn's historical article Localization in general rings, a historical survey - as well as other papers in that volume: Ranicki, A.(ed). Noncommutative localization in algebra and topology. ICMS 2002.
Perhaps the comment refers to the fact that in order to generalize rings to structures with noncommutative addition, we cannot simply delete the axiom that addition is commutative, since, in fact, other axioms force addition to be commutative (Hankel, 1867 [1]). The proof is simple: apply both the left and right
distributive law in different order to the term $\rm\:(1\!+\!1)(x\!+\!y),\:$ viz.
$$\rm (1\!+\!1)(x\!+\!y) = \bigg\lbrace \begin{eqnarray}\rm (1\!+\!1)x\!+\!(1\!+\!1)y\, =\, x + \color{#C00}{x\!+\!y} + y\\
\rm 1(x\!+\!y)\!+1(x\!+\!y)\, =\, x + \color{#0A0}{y\!+\!x} + y\end{eqnarray}\bigg\rbrace\Rightarrow\, \color{#C00}{x\!+\!y}\,=\,\color{#0A0}{y\!+\!x}\ \ by\ \ cancel\ \ x,y$$
Thus commutativity of addition, $\rm\:x+y = y+x,\:$ is implied by these axioms:
$(1)\ \ *\,$ distributes over $\rm\,+\!:\ \ x(y+z)\, =\, xy+xz,\ \ (y+z)x\, =\, yx+zx$
$(2)\ \, +\,$ is cancellative: $\rm\ \ x+y\, =\, x+z\:\Rightarrow\: y=z,\ \ y+x\, =\, z+x\:\Rightarrow\: y=z$
$(3)\ \, +\,$ is associative: $\rm\ \ (x+y)+z\, =\, x+(y+z)$
$(4)\ \ *\,$ has a neutral element $\rm\,1\!:\ \ 1x = x$
In order to state this result concisely, recall that a SemiRing is
that generalization of a Ring whose additive structure is relaxed
from a commutative Group to merely a SemiGroup, i.e. here the only
hypothesis on addition is that it be associative (so in SemiRings,
unlike Rings, addition need not be commutative, nor need every
element $\rm\,x\,$ have an additive inverse $\rm\,-x).\,$ Now the above result may
be stated as follows: a semiring with $\,1\,$ and cancellative addition
has commutative addition. Such semirings are simply subsemirings
of rings (as is $\rm\:\Bbb N \subset \Bbb Z)\,$ because any commutative cancellative
semigroup embeds canonically into a commutative group, its group
of differences (in precisely the same way $\rm\,\Bbb Z\,$ is constructed from $\rm\,\Bbb N,\,$
i.e. the additive version of the fraction field construction).
Examples of SemiRings include: $\rm\,\Bbb N;\,$ initial segments of cardinals;
distributive lattices (e.g. subsets of a powerset with operations $\cup$ and $\cap$;
$\rm\,\Bbb R\,$ with + being min or max, and $*$ being addition; semigroup semirings
(e.g. formal power series); formal languages with union, concat; etc.
For a nice survey of SemiRings and SemiFields see [2]. See also Near-Rings.
[1] Gerhard Betsch. On the beginnings and development of near-ring theory.
pp. 1-11 in:
Near-rings and near-fields. Proceedings of the conference
held in Fredericton, New Brunswick, July 18-24, 1993. Edited by Yuen Fong,
Howard E. Bell, Wen-Fong Ke, Gordon Mason and Gunter Pilz.
Mathematics and its Applications, 336. Kluwer Academic Publishers Group,
Dordrecht, 1995. x+278 pp. ISBN: 0-7923-3635-6 Zbl review
[2] Hebisch, Udo; Weinert, Hanns Joachim. Semirings and semifields. $\ $ pp. 425-462 in: Handbook of algebra. Vol. 1. Edited by M. Hazewinkel.
North-Holland Publishing Co., Amsterdam, 1996. xx+915 pp. ISBN: 0-444-82212-7
Zbl review,
AMS review
Best Answer
Hagen von Eitzen gave a noncommutative nonunital ring with this property in the comments. However if your ring has a unit, this is impossible.
The proof below is a mess of algebra, so let me briefly explain the idea.
The problem is that the equalities we are given are homogeneous equations of degree at least $4$, like $abab=a^2b^2$, and we want a homogeneous equation of degree 2, $ab=ba$. Thus we need to somehow dehomogenify these equations, which we can do by substituting $a=1+a$ or $b=1+b$ and seeing what we get. This is the idea of the proof below.
Edit:
I will leave my original proof below, but darij grinberg suggested an excellent way to simplify the calculations by phrasing things in terms of commutators in the comments below.
Observe that the condition $abab=a^2b^2$ for all $a$ and $b$ is the same as saying $$a[a,b]b=0$$ for all $a$ and $b$. Now note that substituting $1+a$ in for $a$ gives $$(a+1)[a+1,b]b=(a+1)[a,b]b=0,$$ since $[1,b]=0$. Similarly, we also get $$a[a,b](b+1)=0\text{, and }(a+1)[a,b](b+1)=0.$$ Now add the first and fourth equations and subtract the middle two to get: $$a[a,b]b - (a[a,b]b + [a,b]b) - (a[a,b]b+a[a,b]) + (a[a,b]b + a[a,b] + [a,b]b + [a,b]) =0,$$
or, on simplifying, $$[a,b]=0,$$ as desired.
Orignal proof
Consider $$a^2+2a^2b+a^2b^2=a^2(1+b)^2$$ $$=(a(1+b))^2=(a+ab)^2 = a^2+a^2b + aba + abab.$$ Now use that $a^2b^2 = abab$, and cancel $a^2+a^2b+abab$ from both sides to get $$ a^2b = aba.$$
By symmetry, we must also have $ba^2=aba$, by considering $(1+b)^2a^2$.
Now consider $(1+a)^2(1+b)^2$. Expanding this directly, we get $$(1+2a+a^2)(1+2b+b^2)$$ $$ = 1+2b+b^2 + 2a+4ab + 2ab^2 + a^2 + 2a^2b + a^2 b^2$$ On the other hand, we also have $$(1+a)^2(1+b)^2=((1+a)(1+b))^2 = (1+a+b+ab)^2,$$ which expands to $$ 1 + a + b + ab + a + a^2 + ab + a^2b + b + ba + b^2 + bab + ab + aba + ab^2 + abab $$ $$ = 1 + 2a + 2b + 3ab + a^2 + a^2b + ba + b^2 + bab + aba + ab^2 + abab. $$ Now use that $bab=ab^2$, $aba=a^2b$, and $abab=a^2b^2$ to rewrite this as $$ 1 + 2a + 2b + 3ab + a^2 + 2a^2b + ba + b^2 + 2ab^2 + a^2b^2. $$ Thus we end up with the equality, $$ 1 + 2a + 2b +4ab + a^2 + b^2 + 2ab^2 + 2a^2b + a^2 b^2$$ $$=1 + 2a + 2b + 3ab + ba + a^2 + b^2 + 2a^2b + 2ab^2 + a^2b^2. $$ Cancelling $1+2a+2b+3ab+a^2+b^2+2a^2b+2ab^2+a^2b^2$, we end up with $$ab=ba,$$ as desired.