My question is the following:
If we take the set theoretic complement of the set of zero divisors in a commutative ring with identity will the resulting set be a field? If not is there a process to take out more elements so that at we arrive at a field.
On
Sambo has given a clear answer. Assuming we modify your definition to put back zero, after removing zero-divisors, we can still find interesting counter-examples which are not trivial.
Take $F$ to be any field (say real numbers, if you prefer). The in the commutative ring $R=F\times F$ the zero divisors are elements of the form $(a,0)$ and $(0,b)$.
The resulting set is closed under multiplication, but not under addition, so not a subgroup: $(2,3) + (-2,3) = (0,3)$ which is a zero-divisor.
If you remove any set including the zero divisors, the result won't be a field (using the same operations + and ×) because you'll no longer have a $0$ element!