If $A=\varnothing$, it doesn’t matter what $B$ is: $A\setminus B=\varnothing\setminus B=\varnothing$, and $A\cap B=\varnothing\cap B=\varnothing$, so $(A\setminus B)\oplus(A\cap B)=\varnothing\oplus\varnothing=\varnothing=A$.
But in fact it’s always true, for any sets $A$ and $B$. If $x\in A\cap B$, then $x\in B$, so $x\notin A\setminus B$. Conversely, if $x\in A\setminus B$, then $x\notin B$, so $x\notin A\cap B$. Thus,
$\begin{align*} (A\setminus B)\oplus(A\cap B)&=\Big((A\setminus B)\setminus(A\cap B)\Big)\cup\Big((A\cap B)\setminus(A\setminus B)\Big)\\ &=(A\setminus B)\cup(A\cap B)\\ &=A\;. \end{align*}$
If you’re in doubt about that last step, notice that $A\setminus B$ consists of the things that are in $A$ but not in $B$, while $A\cap B$ consists of those things that are in both $A$ and $B$, so between them they pick up every element of $A$.
More generally, the symmetric difference of two disjoint sets is always their union: if $X\cap Y=\varnothing$, then $X\oplus Y=X\cup Y$. Here the disjoint sets are $A\setminus B$ and $A\cap B$, and their union is $A$.