Rings of fractions for all possible sets of denominators This is related to a question I asked here except in this case, I am not asking about the total (complete) ring of fractions. I am asking how to find all possible rings of fractions up to isomorphism for all possible denominators.
Specifically, I am asking for the cases when the rings in question are

*

*$R = \mathbb{Z}_{p}$, where $p$ is a prime. (Note: This is NOT the p-adic integers. This is the set $\{0, 1, \cdots, p-1 \}$ of integers modulo a prime $p$.)

*$R=\mathbb{Z}_{p^{2}}$, again, where $p$ is a prime.

I know that when $R = \mathbb{Z}_{p}$, which is a field, the only multiplicative subgroups are the zero subgroup, $U(\mathbb{Z}_{p})$ (the units of $\mathbb{Z}_{p}$), and of course $\mathbb{Z}_{p}$ itself (although I don't know how to prove this).
When we are discussing the total ring of fractions, the set $D$ (sometimes called $S$) of denominators consists of all non-zero, non zero divisor elements of $\mathbb{Z}_{p}$. However, here, I am asked to consider ALL possible denominators. When $D$ contains $0$, what I get for my ring of fractions is the zero ring, correct (so this would also include the case when $D$ is the whole ring?)? When $D$ is the set of units, then I have the total ring of fractions, and I have, by the question I linked, that $D^{-1}\mathbb{Z}_{p} \simeq \mathbb{Z}_{p}$? Is any of what I'm reasoning here correct?
But what about thee case when $R = \mathbb{Z}_{p^{2}}$? What are all the possible sets of denominators in this case? Recall that the set of denominators just has to be a sub-semigroup. So maybe even for the case where $\mathbb{Z}_{p}$, I'm not including all the possible sets of denominators!
Please help, I am very confused. Thank you.

Where I'm getting stuck is - I don't know how to decide what all the $D$'s should be that I need to check. Because inversion by units doesn't change anything from the complete (total) ring of fractions, I suppose I should just consider $D$'s containing zero divisors and a $D$ consisting of just $0$ itself? Then, how do I figure out what they're isomorphic to?
 A: *

*$R = \mathbb{Z}_{p}$, $p$ prime:



First, note that since $p$ is prime, $\mathbb{Z}_{p}$ is a field;
  i.e., $\mathbb{Z}_{p}$ has no zero divisors. I.e., $\forall m \in
\mathbb{Z}_{p}$ where $m \neq 0$, $\exists n \in \mathbb{Z}_{p}$such
  that $mn = nm = 1$ -i.e., every nonzero element of $\mathbb{Z}_{p}$ is
  a unit.
Now, suppose $D$ consists only of units, then $D \subseteq U(
 \mathbb{Z}_{p})$. Also, note that $\mathbb{Z}_{p}$ has the identity. So, we apply the following result: 

Proposition - Let $R$ be a commutative ring and $D$ be a nonempty multiplicative subsemigroup of $R$. Then the following is true:


*

*The map $\phi: R \to D^{-1}R$ defined by 
$$ \phi(x) = \frac{xy}{y}\, \text{for some}\,y \in D. $$
is a ring homomorphism.

*If $0 \notin D$ and $D$ has no zero divisors, then $\phi$ is a monomorphism.

*The image $\phi(x)$ of every element $x \in D$ is invertible; i.e., $\phi (D) \subseteq U(D^{-1}R)$.

*If $R$ has the identity and $D \subseteq U(R)$ then $\phi$ is an isomorphism.



to conclude that the map $\displaystyle \phi: \mathbb{Z}_{p}\to D^{-1}\mathbb{Z}_{p}$ defined by $\displaystyle \phi(x) = \frac{xy}{y}$ for some $y \in D$ is an isomorphism. Thus, in the case when $D$ consists only of units, the ring of fractions is $D^{-1}\mathbb{Z}_{p} \simeq \mathbb{Z}_{p}$. 
Second, suppose instead that $0 \in D$. In order for the pairs $(x,y)$, $(x^{\prime},y^{\prime}) \in R \times D$ to be equal, there must exist some $z \in D$ such that 
  $$ (xy^{\prime}-yx^{\prime})z = 0.$$
Now, according to my notes, this guarantees that $D^{-1}R$ must be the
  zero ring, but I am not sure why in terms of $(xy^{\prime} -
yx^{\prime})z=0$ that is true. If someone could explain that to me
  either in the comments or in a different answer, I would be most
  appreciative.
So, to summarize, in the case that $0 \in D$, $D^{-1}\mathbb{Z}_{p} =
 \{0\}$.



*

*$R = \mathbb{Z}_{p^{2}}$, $p$ prime:



In the case that $R = \mathbb{Z}_{p^{2}}$, the set of units is the set
  of all $m \in \mathbb{Z}_{p^{2}}$ such that $\gcd(m, p^{2}) = 1$.
So, suppose first that $D$ consists only of units of
  $\mathbb{Z}_{p^{2}}$. Then, $D \subseteq U(\mathbb{Z}_{p^{2}})$, and
  we note that $\mathbb{Z}_{p^{2}}$ has the identity. Therefore, by the
  proposition given in the first bullet point, the map $\phi:
\mathbb{Z}_{p^{2}} \to D^{-1}\mathbb{Z}_{p^{2}}$ defined by
  $\displaystyle \phi(x)=\frac{xy}{y}$ for some $y \in D$ is an
  isomorphism. Thus, in the case when $D$ consists only of units, the
  ring of fractions is $D^{-1}\mathbb{Z}_{p^{2}} \simeq
\mathbb{Z}_{p^{2}}$.
Second, suppose that $0 \in D$. By the remarks in the first bullet
  point, again, this is the case where $D^{-1}\mathbb{Z}_{p^{2}}$ is the
  zero ring.
Third, suppose that $D$ contains a zero divisor. Now, zero divisors of
  $\mathbb{Z}_{p^{2}}$ are multiples of $p$.
Since $D$ is a multiplicative semisubgroup, if $ap \in D$ where $ap$
  is a multiple of $p$, then $D$ contains $(ap)\cdot (ap) = (ap)^{2} =
a^{2}p^{2} = a^{2} \cdot 0 = 0$.
So, if $D$ contains a zero divisor, it also contains $0$ itself, which
  degenerates to the case where $0 \in D$; i.e., in this case also,
  $D^{-1}\mathbb{Z}_{p^{2}}$ is the zero ring.

