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.
Measure Theory using $\sigma$-rings will lead to more complex notion of measurable function, with some non-intuitive results.
Let $\Omega$ be a set and let $\Sigma$ be a $\sigma$-algebra. Let $f$ be a function from $\Omega$ to $\mathbb{R}$. We say that $f$ is measurable if for every Borel set $B$ in $\mathbb{R}$, $f^{-1}(B)\in \Sigma$.
Now, suppose that $\Sigma$ is a $\sigma$-ring and we try to use the same definition. Then, since $\Omega=f^{-1}(\mathbb{R})$, either $\Sigma$ is a $\sigma$-algebra or there will be no measurable function from $(\Omega,\Sigma)$.
So, when working with $\sigma$-rings, we need a slightly different definition (as we find in Halmos' book). We say that $f$ is measurable if for every Borel set $B$ in $\mathbb{R}$, $[f\neq 0]\cap f^{-1}(B)\in \Sigma$.
This second definition allows the existence of measurable functions even if $\Sigma$ is just a $\sigma$-ring and not a $\sigma$-algebra. However, it leads to a few non-intuitive results. For instance: assume $\Sigma$ is just a $\sigma$-ring and not a $\sigma$-algebra. Then any non-zero constant function is NOT mensurable. As a consequence, if $f$ is measurable, then it is easy to prove, for instance, that $f+1$ is NOT measurable.
So, the theory of measurable and integrable functions is more naturally developed by using $\sigma$-algebras, instead of just $\sigma$-rings.
Best Answer
A field of sets is a family $\mathcal F$ of subsets of a given set $X$ satisfying the axioms:
In other words, it's a boolean algebra of sets with the usual operations. Algebra, in this context, is actually synonymous to field. A $\sigma$-field (-algebra) corresponds to a $\sigma$-complete boolean algebra.
Worth mentioning, it actually is quite naturally a ring in the usual algebraic sense (like any boolean algebra). You're right that it can't be a field except the most trivial two-element case (as zero divisors abound).