Equivalence of Definitions of Integral Domain

$(1)$ implies $(2)$
Let $\struct {D, +, \circ}$ be an integral domain in sense $1$.

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

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

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

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

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

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

$(2)$ implies $(1)$
Let $\struct {D, +, \circ}$ be an integral domain in sense $2$.

$\struct {D, +, \circ}$ is already a commutative ring.

Furthermore, as $\struct {D^*, \circ}$ is a monoid, it is nonempty.

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

It remains to show that $\struct {D, +, \circ}$ has no zero divisors.

We know all elements of $\struct {D^*, \circ}$ are cancellable.

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