If by "is it correct" you mean "is it equivalent to the normal definition" (closed under pairwise intersections, pairwise unions, contains $\emptyset$ and $A$) then the answer is "yes."
The second bullet asserts it is closed under complements and the third under intersections.
Suppose $S,T\in B$. Since $S\cup T=A\setminus((A\setminus S)\cap (A\setminus T))$, and the right hand side consists of complements and intersections of things in $B$, $S\cup T$ is in $B$ as well.
Finally, $A\setminus\emptyset=A$ is in $B$ also, so you have both identities.