Equivalence of Definitions of Integral Domain

$(1)$ implies $(2)$
Assume $\left({D, +, \circ}\right)$ is an integral domain in sense $1$.

As $\left({D, +, \circ}\right)$ is already a commutative ring, it remains to show that $\left({D^*, \circ}\right)$ is a monoid.

Because $\circ$ is a ring product, and $\left({D, +, \circ}\right)$ has no zero divisors, we conclude Closure and Associativity.

Furthermore, $\left({D, +, \circ}\right)$ is non-null, hence $0_D \ne 1_D$, and we conclude $1_D \in D^*$.

Therefore, we also have an Identity for $\left({D^*, \circ}\right)$, and hence it is a monoid.

It remains to show that all elements of $\left({D^*, \circ}\right)$ are cancellable.

As $\left({D, +, \circ}\right)$ has no zero divisors, this follows from Ring Element is Zero Divisor iff not Cancellable.

$(2)$ implies $(1)$
Assume $\left({D, +, \circ}\right)$ is an integral domain in sense $2$.

$\left({D, +, \circ}\right)$ is already a commutative ring.

Furthermore, as $\left({D^*, \circ}\right)$ is a monoid, it is nonempty.

Also, we conclude that $\left({D, +, \circ}\right)$ is a non-null ring with unity.

It remains to show that $\left({D, +, \circ}\right)$ has no zero divisors.

We know all elements of $\left({D^*, \circ}\right)$ are cancellable.

From Ring Element is Zero Divisor iff not Cancellable, we conclude that $\left({D, +, \circ}\right)$ cannot have zero divisors.