Let $X$ be a given set and let $A$ be its subset. Define $D$ as a map from the power set of $X$ to itself such that $D(B)=(A \setminus B) \cup (B\setminus A)$. I have already proved that $D$ is injective and I can't seem to conclude if the following statements are true (and if not, one needs to provide a counter example)
Let $A \triangle B$ denote the symmetric difference of $A$ and $B$. You should convince yourself that $A\triangle \emptyset=A$ and $A \triangle A =\emptyset$, that $A\triangle B= B \triangle A$ and that $A\triangle (B\triangle C)=(A\triangle B) \triangle C$. With these properties, it is clear that $C = A \triangle (A \triangle C)$ for any sets $A$ and $C$. Hence, in the language of your question, for a fixed $A$, Let $D(B)=A\triangle B$. To show that $D$ is surjective, let $C$ be an arbitrary subset of $X$. Then $D(D(C))=C,$ so every $C$ is in the image of $D$. Bijectivity is just injectivity together with surjectivity.
