Define $Q(R,T)$ as follow:
$$
(a,b)\in Q(R,T)\iff a\in R,b\in T,\: b\neq0\tag1
$$
For any $a,c\in R, b,d\in T,\:b\neq0,d\neq0$ ($T$ contains no zero divisor of $R$)
$$
(a,b)=(c,d)\iff ad=bc\tag2
$$
$$
(a,b)+(c,d)=(ad+bc, bd)\tag3
$$
$$
(a,b)\cdot(c,d)=(ac, bd)\tag4
$$
Since $T$ contains no zero divisor of $R$, $bd\neq0$ as long as $b\neq0$ and $d\neq0$. So $(3)$ and $(4)$ always hold.
We can prove $Q(R,T)$ be a commutative ring with unity, where $(0,a)$ is zero and $(a,a)$ is unity ($a\in T,\:a\neq0$), by proving all ring axioms hold for $Q(R,T)$. For example,
- Commutativity of addition:
$$
(a,b)+(c,d)=(ad+bc, bd)=(cb+da, db)=(c,d)+(a,b)
$$
- Commutativity of multiplication:
$$
(a,b)\cdot(c,d)=(ac, bd)=(ca,db)=(c,d)\cdot(a,b)
$$
- Additive zero:
$$
(a,b)+(0,d)=(ad+b0,bd)=(ad,bd)=(a,b)
$$
- Multiplication unity:
$$
(a,b)\cdot(d,d) = (ad,bd)=(a,b)
$$
and rest follow.
In addition, for any $a\neq0,\:b\neq0$
$$
(a,b)\cdot(b,a)=(ab,ba)=(c,c)
$$
Thus every nonzero element of $T$ is a unit.
Edit:
To prove that $R$ can be enlarged to $Q(R,T)$, consider $Q(R,a), \:a\in T$. Clearly
$$
Q(R,a)\cong R\quad\text{and}\quad Q(R,a)\subset Q(R,T)
$$
Thus $R$ can be enlarged to a partial ring of quotients $Q(R,T)$.
Best Answer
You do not need commutativity to define a unit. However, the multiplicative inverse of an element must necessarily commute with that element. That is, if $ u \in R $ is a unit and $ v $ is its inverse, that by definition means $ uv = 1 = vu $. This is the same as the defintion of the inverse in a (not necessarily abelian) group.