Does the cancellation property for a group mean something different than the cancellation property for an integral domain? I recently started learning about rings and some of their elementary characteristics / basic properties. One of the concepts that sort of caught me off guard was the statement that infinite ordered integral domains are not necessarily fields. I thought about it and saw that $\mathbb Z$ was a concrete example of this because other than the elements $1, -1$, no other elements have multiplicative inverses. 
Integral domains are defined as: a commutative ring with unity having the cancellation property $\iff$ commutative ring with unity having no divisors of $0$
Fields are defined as: a commutative ring with unity in which every nonzero element is invertible
From prior learning about groups, the proof that groups exhibit the cancellation product employed a strategy that invoked inverse elements. (i.e. $ax=bx \implies axx^{-1}=bxx^{-1} \implies a=b$)
If a particular integral domain is not a field (and therefore exhibits the cancellation property but not all elements have multiplicative inverses), does that mean that the cancellation property of some integral domains is fundamentally different than the cancellation property of a group? 
I ask this because the proof strategy for demonstrating that such an integral domain exhibits the cancellation property must be fundamentally different from the strategy that is used in the group-proof (because invertible elements are generally absent). 
 A: If you look at the case of the integers $\mathbb Z$, you can prove that it has the cancellation property by using the fact that it has no zero divisors. However, many people would take an approach that is technically much more difficult by arguing that $\mathbb Z$ can be embedded (as a ring) in the field of rational numbers $\mathbb Q$, in which the cancellation property holds because of the existence of inverses. This other argument can also be extended to arbitrary integral domains, by considering the field of fractions of the integral domain. 
Whether this makes the cancellation property fundamentally the same for integral domains and fields, I'll leave for you to judge. I think the important point is that, even if such constructions are possible, they are by no means needed to prove the cancellation property.
A: In order to prove that, in a commutative ring $(R,+\times)$, the cancellation property holds, you cannot assume that every non-zero element has an inverse; you are not assuming that $(R\setminus\{0\},\times)$ is a group (if it was, your commutative ring would be a field.
For instance, $\mathbb Q[x]$ is an integral domain in checking this means, in particular, that you should check that$$P(x),Q(x)\in\mathbb Q[x]\setminus\{0\}\implies P(x)Q(x)\neq0.$$
