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,
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.