To show that $P$ is a subring, we need to show that it is a ring. That is: we need to show that:
- Addition in $P$ is associative.
- Addition in $P$ is commutative.
- Multiplication in $P$ is associative.
- Multiplication is left and right distributive over addition.
- For all $x, y \in P$, there is $x + y \in P$.
- For all $x, y \in P$, there is $xy \in P$.
- $0 \in P$
- For all $x \in P$, there is $-x \in P$.
- $1 \in P$ (see note).
Note: opinions vary on whether rings must have multiplicative identities. I like them to, so I've included it.
I've listed them in that slightly odd order for a reason: the first four are all immediate: they're all true in all of $\mathbb{R}$, and don't have any existential quantification going on, so they're true in $P$ as well.
So we actually only have those last five things to check, and we can, if we like, just go through and check them all separately. However, we can be lazy:
If $P$ is closed under subtraction and nonempty, then we have some $x \in P$, so we have $0 = x - x \in P$ (so we don't need to check $0 \in P$ separately), and also $0 - x = -x \in P$ (so we don't need to check that $P$ has additive inverses separately), and finally for any $y \in P$, we have $y - (-x) = y + x = x + y \in P$ (so we don't need to check closure under addition separately. However, $0$ is very often the easiest thing to show that $P$ contains, so very often we'll do that, then subtraction, and have all of the addition axioms sorted.
So to finish the job off, we only need to check that $1 \in P$ (if we like our rings to have identities) and $xy \in P$ for all $x, y \in P$.
However, we can be even lazier: we can look ahead, and check that it's a subfield, which will then immediately imply that it's a subring.
We could check that it's a subfield by just doing the above, then checking the extra field axioms (commutativity of multiplication is immediate as with the first four above, we now definitely need $1 \in P$ if we didn't already, we need to check multiplicative inverses, and we need $1 \neq 0$, but that's also immediate from $\mathbb{R}$).
We could, again, just check both of the new things that need checking. However, we can now do the same thing as we did with addition above, and just check that $P$ is closed under non-zero division and that $P$ contains something other than $0$, and get the rest for free (we then have $x \in P \setminus \{0\}$, and so $1 = xx^{-1}$ and $x^{-1} = 1x^{-1}$, and finally $xy = y(x^{-1})^{-1}$ for all $y \in P$).