No, you can't get complements from unions and intersections. For example, let $X$ be a nonempty set. Then $\{\emptyset\}$ is nonempty, closed under (arbitrary) intersection and union, but not closed under complements.
You can get intersections from unions and complements using De Morgan's laws. To get an intersection, just take the complement of the union of the complements. Similarly, you get unions from intersections and complements. So yes, the definition of field of subsets was redundant.
I think there is no non-trivial way to do what the question is asking.
Let's say that a set of partial (possibly total) binary operations $S$ is complete for the symmetric difference operation $\triangle$ if
every operation in $S$ is a (possibly trivial) restriction of some composition of copies of $\triangle$, and
for every pair of sets $(A,B)$ there is a partial operation $\cdot$ that is a composition of partial operations in $S$ and has the property that $A \cdot B$ is defined and is equal to $A \mathbin{\triangle} B$. (Note that is weaker than saying that $S$ itself is a composition of partial operations in $S$, which is impossible if none of the operations in $S$ is total.)
Even with this weak version of the definition, the only sets $S$ that are complete for $\triangle$ are trivial, in a sense that we will make precise below. Note that by the first clause of the definition, and the fact that $\triangle$ is commutative, associative, and satisfies $A \mathbin{\triangle} A = \emptyset$, every partial operation $\cdot$ in $S$ is the restriction of one of the four following total operations:
The constant binary operation given by $A \cdot B = \emptyset$,
The projection given by $A \cdot B = A$,
The projection given by $A \cdot B = B$, and
The symmetric difference operation $\triangle$ itself.
The first three operations do not help us form any additional binary operations under composition, so we consider them to be trivial, and we assume that $S$ consists only of restrictions of $\triangle$.
A trivial way to get a complete set $S$ is for the union of the domains of the partial operations in $S$ to consist of all pairs $(A,B)$: for example, the proper set difference operation "$\setminus$" as defined in the question is simply the restriction of $\triangle$ to the domain $\{(A,B) : B \subseteq A\}$, so we could let $S = \{\setminus, \cdot\}$ where $\cdot$ is the restriction of $\triangle$ to the complementary domain $\{(A,B) : B \not \subseteq A\}$. More generally, we could weaken the triviality requirement to consider the domains of the partial operations in $S$ and their reverses; for example, this requirement would be satisfied by the set $S = \{\setminus, \cdot\}$ where $\cdot$ now denotes the restriction of $\triangle$ to the domain $\{(A,B) : A \not \subseteq B\}$.
However, every complete set is trivial in this way. Suppose that some pair $(A,B)$ is not in the domain of any partial operation $\cdot$ in $S$, and neither is the reversed pair $(B,A)$. Because the partial operations in $S$ are all binary, it is not hard to show that the pairs $(A,B)$ and $(B,A)$ cannot be in the domain of any composition of elements of $S$ either, so there is no way to get the symmetric difference $A \mathbin{\triangle} B$.
Best Answer
There is a "minimal" definition of set operations, though it's not really a restriction of the classical ones, and is more a fun trick than a clever insight. It was discovered by Charles Sanders Peirce, for general Boolean algebras.
Let $A\uparrow B=(A\cap B)^c$. Then $A\uparrow A=A^c$. So $(A\uparrow A)\uparrow(B\uparrow B)=(A^c\cap B^c)^c=A\cup B$. Similarly, $(A\uparrow B)\uparrow(A\uparrow B)=((A\cap B)^c)^c=A\cap B$. And then you can get difference or whatever else, as expected. You can do the same thing with $(A\cup B)^c$.