I am currently studying integral domains and see that if $ab = ac$ and $a \neq 0$, then we have $b = c$. I was wondering whether the same is true for three nonzero subsets $I, J, K$ of an integral domain $R$, namely, if $I, J, K$ be three nonzero subsets of an integral domain $R$ such that none of them are equal with $R$ and contains $1$ and satisfies $IJ = IK$, then does this imply that $J = K$? (For nonempty subsets $A, B$ of $R$, we mean $AB = \{ab \mid a \in A \text{ and } b \in B\}$).
Also, if the answer is not affirmative, then what kind of domains satisfy this cancellative property about subsets?
The real numbers are a field, and therefore an integral domain, and yet $$ (0,\infty)\cdot (1,\infty)=(0,\infty)\cdot (0,\infty)=(0,\infty) \, . $$ In fact, any integral domain $R$ which satisfies the desired property is isomorphic to $\mathbb Z/2\mathbb Z$. The proof is simple: since $$ \left(R\setminus\{0\}\right)\cdot \left(R\setminus \{0\}\right)=\left(R\setminus\{0\}\right)\cdot \{1\}=R\setminus\{0\} \, , $$ it follows that $R\setminus \{0\}=\{1\}$, so $R=\{0,1\}$. Since $1\neq0$ in an integral domain, $R$ has two elements, and any ring with two elements is isomorphic to $\mathbb Z/2\mathbb Z$. (Note that we also implicitly used the fact that $1\neq0$ to conclude that $R\setminus\{0\}$ contains $1$ when we were doing the above calculation.)
Even in $\mathbb Z/2\mathbb Z$, we have $\{0\}\cdot\{0\}=\{0\}\cdot \{1\}$, but I understand from your question that this is okay.