I was wondering why an integral domain is required to be commutative and have a multiplicative neutral element.
For example the quaternions would be a non-commutative integral domain.
The interesting property is that it has no zero divisors, so why require it to be commutative and have a $1$?
In other words: what consequences follow from having a one vs. not having one when investigating rings without zero divisors?
Yes, people do study rings without unity (see for example this previous answer for some comments on rings without unity). So people do study commutative-rings-without-unit-that-have-no-zero-divisors.
Likewise, people do study non-commutative rings (with or without identity), and study such rings that have no zero divisors (though from what I can tell from my colleagues, zero divisors don't worry them quite as much as nilpotent elements do, so "reduced rings" seem to play a bigger role). So certainly, people study things that are "almost domains except for one of the properties" (nobody drops the $1\neq 0$ condition, though, because that makes it too easy).
A ring with 1 which is not the zero ring and has no zero divisors is called a domain (at least in Lam's First Course in Noncommutative Rings). They are defined on page 3 of the book, so pretty early. They can get pretty hairy: if you have a domain, it may not be embeddable in a division ring (in contrast to the integral domain case, where you always have the field of fractions); this was proven by Mal'cev, and is presented in Lam's Lectures on Rings and Modules, section 9B.
So really your question is more along the lines of "Why do we distinguish precisely integral domains, and not some of these other objects?" Maybe we should study rings without zero divisors, give them a special name (say, "wuzeds", for "without zero divisors", and then call integral domains "commutative wuzeds"?)
Partly, history. Ring theory was born from ideas of Dedekind, and later Noether and Artin, which arose from number theory considerations; they all took place in the commutative setting, and most of what Dedekind worked with were in fact (a special family of examples of) what we now call "integral domains". Integral domains were simply studied more. Also, a lot of ring theory originally arose as a way to try to abstract these ideas, generalize the results; a lot of the results were based on polynomials, on matrices, and other objects constructed from the rings, which people were familiar with in the case of the integers and other prototypes. Most of these objects become very hard when you switch to rings without identity (for example, whereas every ring with identity can be embedded in both its polynomial and its matrix rings in canonical ways, rings without identity are far more difficult to embed), or when you switch to noncommutative rings.
So people were studying special cases of commutative rings before they were studying noncommutative rings. That's also why we call them fields and skew-fields (or division rings), and not "commutative division rings" and "division rings". (Bourbaki attempted to change this, by defining "field" to be a ring with identity, $1\neq 0$, in which every nonzero element has an inverse, i.e. what we usually call a division ring, and then talking about fields and commutative fields; it did not stick outside of France).
Why do you say that the interesting property of an integral domain is that it not have zero divisors? This is giving short shrift to the other properties, i.e. having 1, and being commutative.
The question of rings without 1, and their relevance within mathematics, has been discussed in various places here and on MO: see e.g. here, here, and here.
The differences between commutative and non-commutative rings are immense. Essentially, in commutative rings one can localize, and in non-commutative rings one cannot. See e.g. this answer for more discussion of this.
The theory of integral domains is based on all three properties (commutativity, existence of 1, no zero-divisors).
To give a concrete example, the enveloping algebra of a Lie algebra over a field has no zero divisors. If the Lie algebra is of finite dimension, it is Noetherian. (One sees this because the enveloping algebra admits a filtration, the filtration by degree, whose associated graded is a polynomial ring --- so in particular a domain --- in as many variables as the dimension of the Lie algebra --- and so Noetherian for a finite dimensional Lie algebra.)
Now lets suppose our Lie algebra $\mathfrak g$ is finite dimensional and semi-simple over a field of characteristic zero, say $\mathbb C$ just to fix ideas. The enveoping algebra $U(\mathfrak g)$ contains an augmentation ideal (the kernel of the action on the trivial representation, if you like), call it $I$; so there is a short exact sequence $$0 \to I \to U(\mathfrak g) \to \mathbb C \to 0.$$ Now because $\mathfrak g$ is semi-simple, any extension of the trivial representation $\mathbb C$ by itself splits, and so is again trivial. This implies that $I^2 = I$.
On the other hand, in a Noetherian integral domain (commutative!), one has the result that $I^2 = I$ implies $I = 0$. (See e.g. this answer.)
Geometrically, the idea is that if $A$ is a commutative ring with one and $I$ is an ideal such that $I = I^2$, then Spec $A/I$ is a closed subscheme of Spec $A$ which has no non-trivial normal directions into Spec $A$. If one pictures this, you will see that the only way this seems possible, intuitively, is if Spec $A/I$ is open as well as closed in Spec $A$ (otherwise there would be some directions pointing out of Spec $A/I$ into the rest of Spec $A$). One can then actually prove that if $A$ is furthermore Noetherian, and if $I = I^2$, then $A$ factors as a product $A/I \times B$, so Spec $A$ is the disjoint union of Spec $A/I$ and Spec $B$. In particular, if $A$ is a domain (so that Spec $A$ is irreducible, and so connected), one finds that either $I = A$ or $I = 0$.
The example of the enveloping algebra shows that the non-commutative situation is completely different: we have the ring $U(\mathfrak g)$ without zero divisors, and Noetherian, but with a non-trivial idempotent ideal. This is (at least for me) quite hard to interpret geometrically, and certainly suggests that it is reasonable to separate the study of (commutative) integral domains from the more general study of (possibly non-commutative) rings without zero divisors.
Final remark: I have no objection to using the term integral domain or domain in the non-commutative context. (I should note that, like many commutative algebraists, I frequently say and write domain to mean commutative integral domain.) My answer here is not intended to advocate any position on terminology, but just to explain why the commutative context is quite different from the more general non-commutative one, and hence why it makes sense to make a special study of the commutative case.
Incidentally, my answer could reasonably be taken in a more general sense as (at least partly) explaining why we make a special study of commutative rings (the subject of commutative algebra) rather than just studying all rings at once.
From one general algebraic perspective, the "point" of defining integral domains is really to define their fraction fields. Integral domains are precisely the (unital) subrings of (commutative) fields, so in that sense the study of integral domains is a natural extension of the study of fields. This is why the basic objects in, say, algebraic number theory are certain integral domains which are subrings of number fields.
From the perspective of number theory, the "point" of defining integral domains is to impose an obvious requirement on any ring with a nice theory of factorization: if $0$ can have a nontrivial factorization, then there are going to be problems. Again, this is why in algebraic number theory we study integral domains.
From the perspective of algebraic geometry, the point of defining integral domains is as follows. Associated to any nice integral domain $R$ is a certain topological space, its spectrum, and one should roughly speaking think of elements of $R$ as functions on this space $\text{Spec } R$. Then the requirement that $R$ is an integral domain translates into requiring a very strong connectivity hypothesis on this space. If $a, b \in R$ such that $ab = 0$, then the set of points in $\text{Spec } R$ such that $a = 0$ and the set of points in $\text{Spec } R$ such that $b = 0$ are closed subsets of $\text{Spec } R$ whose union is all of $\text{Spec } R$, and under certain hypotheses on $R$ these subsets are proper. Requiring that $R$ has no zero divisors is (again, under certain hypotheses) equivalent to requiring that it is not possible to partition $\text{Spec } R$ in this way. This is a very strong form of connectivity called irreducibility. It turns out that any (nice) variety is a unique disjoint union of a finite number of irreducible components, so it's natural to study irreducible varieties.
As for requiring units, all of the natural examples I can think of have units, and all of the natural morphisms I can think of preserve them. The category of rings without units is messy and, while it has its place, I don't think it's appropriate for any of the applications I've described above.
Geometrically, consider the following basic example: if $X$ is a topological space, then there is a ring $C(X)$ of continuous functions $X \to \mathbb{R}$. This ring is commutative, and it has a unit, although it is not usually an integral domain. In any case, the idea here is that we can study topological properties of $X$ by studying algebraic properties of $C(X)$, and in particular any continuous map $X \to Y$ gives an $\mathbb{R}$-algebra homomorphism $C(Y) \to C(X)$ in the other direction by precomposition. This ring homomorphism automatically preserves units, so it's not natural to ignore the units here. If $X$ and $Y$ are compact Hausdorff, something amazing happens: all $\mathbb{R}$-algebra homomorphisms $C(Y) \to C(X)$ preserving units come from continuous maps $X \to Y$, so one really can get all the topological properties of compact Hausdorff spaces using commutative unital rings.
There are two things that come to my mind. One thing is that the polynomial ring isn't as nice. For example over a field, the number of roots of a polynomial is bounded by the degree. This isn't the case over a skew-field (a ring satisfying all the axioms of a field except for commutativity). For example, in the quaternions, the innocent looking polynomial $x^2 + 1$ has infinitely many roots (any purely imaginary quaternion with length 1)!
Another thing that gets tricky is linear algebra over skew-fields. Here you have to distinguish between left actions and right actions. Also nice and useful formulas don't hold, e.g. it is not the case that $tr(AB) = tr(BA)$ or that $\det(AB) = \det A \det B$.
Actually you may miss the original base idea of cancellation law. its like this if you have unity element i.e. multiplicative inverse
Then you can do like this ab=ac implies (a^-1)ab=(a^-1)ac implies 1b=1c Because (a^-1)*a=1 Means if you not have unity element then there is a no multiplicative inverse of any element and no cancellation law define without it for multiplication. And for other problem which you have is why its needed commutative. you are actually right about it.
Actually if we need to prove cancellation law means we need to prove right and left both cancellation law.
But on a particular example you may not need both cancellation law so you can you use anyone of those which you required. i.e. if you use left cancellation law like we use in above example then you need to use left identity.
