Ring of Fractions Let $A$ be an integral domain. If $S=A-\left\{0\right\}$, then $S^{-1}A$ is the field of fractions of $A$. 
What is the problem if we actually take $S=A$? From what i see, in that case $0/0=1/1$ and the zero of the ring $S^{-1}A$ will be equal to its identity, thus $S^{-1}A=0$. Is there any other issue of non well-posedness? 
 A: Nothing horrible happens... except that you get the $0$ ring.
Recall that if $S$ is a nonempty multiplicative subset of a commutative ring (not necessarily with $1$) $A$, then $S^{-1}A$ is the ring whose underlying set is the quotient
$$\left.\left\{\frac{a}{s} \, \right|\, a\in A, s\in S\right\}\Bigm/\sim$$
where $\frac{a}{s}\sim \frac{b}{s'}$ if and only if there exists $s''\in S$ such that $s''(as'-bs) = 0$ in $A$; and with addition and multiplication defined by
$$\begin{align*}
\frac{a}{s} + \frac{b}{t} &= \frac{at+bs}{st}\\
\frac{a}{s}\times\frac{b}{t} &= \frac{ab}{st}.
\end{align*}$$
These operations are well defined, make $S^{-1}A$ into a ring with unity (the unity being the class of $\frac{s}{s}$ for any $s\in S$), and there is a natural homomorphism $\varphi\colon A\to S^{-1}A$ given by $\varphi(a) = \frac{as}{s}$, where $s\in S$ is an arbitrary element (this is also well-defined).
There is absolutely no problem if you include $0$ in your multiplicative subset $S$... but if you do, all you get is the zero ring. Because if $0\in S$, then for all $a,b\in A$, $s,s'\in S$, we have $\frac{a}{s}\sim\frac{b}{s'}$, since $0(as'-bs) = 0$ holds. Thus, $S^{-1}A$ has a single element, and so is the $0$ ring.
Because this occurs if you have $0\in S$, many authors exclude the case in which $0\in S$; not because the universe explodes if you put $0\in S$ or anything like that, but merely because as soon as $0\in S$, you just descend into the triviality of the zero ring. This is true with any commutative ring $A$, whether or not it is a domain. 
You can extend the notion of "ring of [left/right] fractions" along similar lines with noncommutative rings, but only in some cases; one large class was studied by Ore, and you can find a lot of results on this in Lam's Lectures on Rings and Modules, Chapter 4. Again, one usually excludes the case of $0\in S$ to avoid trivialities.
