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).
Your reasoning is not wrong. You have shown localization is not injective in general.
Suppose you're localizing at a subset $S\subset A$. You have shown that if $s_1,s_2\in S$ satisfy $s_1s_2=0$ then $0=1\in A[S^{-1}]$. If your $S$ is multiplicatively closed, then your assumptions means $0\in S$, so it may be unsurprising that inverting zero kills the ring.
It can also happen that $s\in S$ is a zero divisor in $A$, but that $s$ does not kill any elements in $S$. In this case, the localization can fail to be injective without being zero. A simple example was given in the comments by user26857: if you localize the quotient ring $\mathbb Z/6\mathbb Z$ at the zero divisor $\hat 2$ you get the field $\mathbb Z/3\mathbb Z$.
Best Answer
Let $R$ be a unital ring, with unit element $1_R$, and let $R'$ be an integral domain (which is not a priori assumed to be unital). Suppose we have a nonzero homomorphism of (nonunital) rings $\phi:R\to R'$. We want to prove that $R'$ is actually unital, with unit element $\phi(1_R)$.
First, we have the following equality: $$\phi(1_R)=\phi(1_R1_R)=\phi(1_R)\phi(1_R).$$ Hence, we see that $\phi(1_R)$ is an idempotent of $R'$. Now, since $\phi$ is nonzero, $\phi(1_R)\neq 0$. Therefore, for any element $x\in R'$, we have $$\phi(1_R)x=(\phi(1_R))^2x \Longrightarrow x=\phi(1_R)x,$$ and similarly, we have $x=x\phi(1_R)$. Therefore, we conclude that $\phi(1_R)$ is a unit element of the ring $R'$.