Let $A$ be a finite commutative ring (not assumed to contain an identity). Suppose that $a \in A$ is not a zero-divisor. Then multiplication by $a$ induces an injection from $A$ to itself, which is necessarily a bijection, since $A$ is finite. Thus multiplication by $a$ is a permutation of the finite set $A$, and hence multiplication by some power of $a$ (which by associativity is the same as some power of multiplication by $a$) is the identity permutation of $A$. That is, some power of $a$ acts as the identity under multiplication, which is to say, it is a (and hence the) multiplicative identity of $A$.
In short, if a finite commutative ring $A$ contains a non-zero divisor, then it necessarily contains an identity, and every non-zero divisor in $A$ is a unit.
The fraction ring (localization) $\rm\,S^{-1} R\,$ is, conceptually, the universal way of adjoining inverses of $\rm\,S\,$ to $\rm\,R.\,$ The simplest way to construct it is $\rm\,S^{-1} R = R[x_i]/(s_i x_i - 1).\,$ This allows one to exploit the universal properties of quotient rings and polynomial rings to quickly construct and derive the basic properties of localizations (avoiding the many tedious verifications always "left for the reader" in the more commonly presented pair approach). For details of this folklore see e.g. the exposition in section 11.1 of Rotman's Advanced Modern Algebra, or Voloch, Rings of fractions the hard way.
Likely Voloch's title is a joke - since said presentation-based method is by far the easiest approach. In fact both Rotman's and Voloch's expositions can be simplified. Namely, the only nonobvious step in this approach is computing the kernel of $\rm\, R\to S^{-1} R,\,$ for which there is a nice trick:
$\quad \begin{eqnarray}\rm n = deg\, f\quad and\quad r &=&\rm (1\!-\!sx)\,f(x) &\Rightarrow&\ \rm f(0) = r\qquad\,\ \ \ via\ \ coef\ x^0 \\
\rm\Rightarrow\ (1\!+\!sx\!+\dots+\!(sx)^n)\, r &=&\rm (1\!-\!(sx)^{n+1})\, f(x) &\Rightarrow&\ \rm f(0)\,s^{n+1}\! = 0\quad via\ \ coef\ x^{n+1} \\
& & &\Rightarrow&\ \rm\quad r\ s^{n+1} = 0
\end{eqnarray}$
Therefore, if $\rm\,s\,$ is not a zero-divisor, then $\rm\,r = 0,\,$ so $\rm\, R\to S^{-1} R\,$ is an injection.
For cultural background, for an outstanding introduction to universal ideas see George Bergman's An Invitation to General Algebra and Universal Constructions.
You might also find illuminating Paul Cohn's historical article Localization in general rings, a historical survey - as well as other papers in that volume: Ranicki, A.(ed). Noncommutative localization in algebra and topology. ICMS 2002.
Best Answer
HINT 1) If $\;\rm R\;$ is finite then $\;\rm x\to r\:x\;$ is onto iff 1-1, so $\;\rm R\;$ is a field iff $\;\rm R\;$ is a domain.
2) is Wedderburn's little theorem.