An element $x/p^k$ of the localization of $\prod_n \mathbb Z/p^n$ is zero if $p^{n-l}$ divides $x_n$ for all $n$ and some $l$. Thus it is a zero divisor if the power of $p$ dividing $x_n$ is unbounded. So a regular element is one where the power of $p$ dividing $x_n$ is bounded, say by $m$. Then there exists some $y$ such that $x_ny_n=p^m$ everywhere, so $xy=p^m$, so $x^{-1}=yp^{k-m}$.
So the localization is classical.(This argument goes trough for any product of Artinian quotients of DVRs.)
Other questions: We can fix this by using a ring of dimension two, say $\prod k[x,y]/(x^n,x^{n-1}y,...,y^n)$. Localize at $y$, then $x$ is no longer a zero divisor because $xf=0$ if and only if $yf=0$, but nothing times $x$ is a power of $y$ so $x$ does not have an inverse.