Identify $S = 0$ for any set $S$ and form a non-boolean ring out of $\Delta$ on sets? Let $S$ be a set and define for each $A \subset S$ the negative of $A$ to be $-A = S \setminus A$.  So that $A + S\setminus A = S$.  Thus we identify $0 = S$.  The $+$ operation is still $\Delta$ so if $-A$ were more than $S \setminus A$ it would be less than $S$ when $\Delta$'d with $A$, similarly it can't be smaller than $S \setminus A$.  Therefore for $|S| \geq 2$, $R$ the ring formed is not boolean, as is the usual case for such $\Delta$ (symmetric difference) rings.  Is this then just $\mathcal{P}(S)/(S)$ where $(S)$ is the ideal generated by the whole set $S \in \mathcal{P}(S)$ the power set of $S$?
Therefore, you can quotient a boolean ring and achieve a non-boolean ring?  That doesn't seem possible, you can't have a surjective $h: A \to B$ such that $A$ is boolean while $B$ is non-boolean.
So where did I make an error?
 A: You do not have any freedom to redefine $-A$ while keeping $+$ the same as it was before.  The zero for $+$ has to satisfy $0+0=0$, and $A\Delta A=A$ implies $A=(A\setminus A)\cup (A\setminus A)=\emptyset$.  So the additive identity is necessarily $\emptyset$, and the additive inverses are necessarily $-A=A$.  Given an operation that makes an abelian group, you are already stuck with the inverses it provides you.
You would have to find a new operation for which $-A$ really behaves as the inverse, in order to succeed.  Unfortunately, this is impossible: since the neutral element is always its own inverse, you would need $Z=Z^c$, which is of course impossible if $S$ is nonempty.  Therefore you don't have an abelian group, much less a ring.
It is true that the quotient of a boolean ring is always a boolean ring.  It is as simple as noting that $x^2=x$ implies $(x+I)^2=x^2+I=x+I$ for all $x$ in the ring.
A: Your structure is not a ring, and not even an Abelian group (since you didn't even specify multiplication).
The only element $x$ in $P(S)$ which satisfies $x\,\Delta\,A=A$ for all $A$ is $x=\emptyset$, so this is the additive identity. (Just plug in $A=\emptyset$.)
Therefore $-A$ must be an element which satisfies $(-A)\,\Delta\,A=\emptyset$, and then $-A=A$ follows.
As an Abelian group, or if you wish, as a vector space over $\Bbb F_2$, your quotient construction can make sense: every $A$ is identified with $S\setminus A=S\,\Delta\,A$, and using associativity you can verify that symmetric difference will work fine in the quotient.
We can even identify it with the Abelian group $(P(S\setminus\{a_0\}),\Delta)$ for a fixed element $a_0\in S$, using the representation $A$ or $S\setminus A$ of a set $A\in P(S)$ which doesn't contain $a_0$.
But, the quotient as rings is the trivial ring $\{0\}$, because $S=1$ is the multiplicative identity (multiplication is intersection), and therefore $1=0$ implies $x=0$ for every element $x$.
