A ring is an ordered triple, $(R,+,\times)$, where $R$ is a set, $+\colon R\times R\to R$ and $\times\colon R\times R\to R$ are binary operations (usually written in in-fix notation) such that:
- $+$ is associative.
- There exists $0\in R$ such that $0+a=a+0=a$ for all $a\in R$.
- For every $a\in R$ there exists $b\in R$ such that $a+b=b+a=0$.
- $+$ is commutative.
- $\times$ is associative.
- $\times$ distributes over $+$ on the left: for all $a,b,c\in R$, $a\times(b+c) = (a\times b)+(a\times c)$.
- $\times$ distributes over $+$ on the right: for all $a,b,c\in R$, $(b+c)\times a = (b\times a)+(c\times a)$.
1-4 tell us that $(R,+)$ is an abelian group. 5 tells us that $(R,\times)$ is a semigroup. 6 and 7 are the two distributive laws that you mention.
We also have the following items:
a. There exists $1\in R$ such that $1\times a = a\times 1 = a$ for all $a\in R$.
b. $1\neq 0$.
c. For every $a\in R$, $a\neq 0$, there exists $b\in R$ such that $a\times b = b\times a = 1$.
d. $\times$ is commutative.
A ring that satisfies (1)-(7)+(a) is said to be a "ring with unity." Clearly, every ring with unity is also a ring; it takes "more" to be a ring with unity than to be a ring.
A ring that satisfies (1)-(7)+(a,b,c) is said to be a division ring. Again, eveyr division ring is a ring, and it takes "more" to be a division ring than to be a ring. (5)+(a)+(b)+(c) tell us that $(R-\{0\},\times)$ is a group (note that we need to remove $0$ because (c) specifies nonzero, and we need (b) to ensure we are left with something).
A ring that satisfies (1)-(7)+(a,b,c,d) is a field. Again, every field is a ring.
We do indeed have that $(R,+)$ is an abelian group, that $(R-\{0\},\times)$ is an abelian group, and that these structures "mesh together" via (6) and (7). In a ring, we have that $(R,+)$ is an abelian group, that $(R,\times)$ is a semigroup (or better yet, a semigroup with $0$), and that the two structures "mesh well".
We have that every field is a division ring, but there are division rings that are not fields (e.g., the quaternions); every division ring is a ring with unity, but there are rings with unity that are not division rings (e.g., the integers if you want commutativity, the $n\times n$ matrices with coefficients in, say, $\mathbb{R}$, $n\gt 1$, if you want noncommutativity); every ring with unity is a ring, but there are rings that are not rings with unity (strictly upper triangular $3\times 3$ matrices with coefficients in $\mathbb{R}$, for instance). So
$$\text{Fields}\subsetneq \text{Division rings}\subsetneq \text{Rings with unity} \subsetneq \text{Rings}$$
and
$$\text{Fields}\subsetneq \text{Commutative rings with unity}\subsetneq \text{Commutative rings}\subsetneq \text{Rings}.$$
If domain means integral domain, then division rings need not be domains because integral domains are commutative. It is conceivable that an unqualified "domain" could mean "a ring with no zero divisors." If that is what the author is using it to mean, then the statement is true: division rings cannot have zero divisors because every nonzero element is invertible.
For a proof, suppose $ab=0$ and $a\neq 0$. Then
$$a^{-1}(ab)=b=0=a^{-1}0$$
Thus whenever $ab=0$, either $a=0$ or $b=0$.
Best Answer
Let $1$ denote the group identity (with respect to multiplication) of $\{ r \in R \mid r\neq 0\}$, which exists by hypothesis. Being an element of $R \setminus \{0\}$, we immediately have $1 \neq 0$. By definition, $1$ is a/the multiplicative identity.