Note: $\bar{x},\bar{z}$ denotes the image of $x$ and $z$ in A and $k$ is a field.
I have a possible proof but I do not know if its correct: Suppose $\bar{a}\bar{b}\in P$ where $a,b\in k[x,y,z]$. Then $\bar{a}\bar{b}=\bar{c}\bar{x}+\bar{d}\bar{z}$ where $c,d\in k[x,y,z]$. Hence, $\overline{ab}=\overline{cx+dz}$. This implies $ab-cx-dz=g(xy-z^2)$ where $g\in k[x,y,z]$. This implies that $ab=cx+dz+g(xy-z^2)$. Therefore, all the monomials involved in the sum on the RHS have either $x$ or $z$ indeterminates or the sum could be zero. If the RHS sum is zero then either $a=0$ or $b=0$. If the RHS sum is not zero then this implies that $a$ and $b$ are both not zero and either $a$ or $b$ only has monomials containing $x$ or $z$. Hence, either $\bar{a}$ or $\bar{b}$ is in P.
Is the proof above valid?
Secondly, I would like to show that $\bar{x}\notin P^2$ and $\bar{y}\notin r(P^2)$. But I'm stuck with that. Any ideas or suggestions?
The ideal $(\bar x , \bar z)$ in the quotient corresponds to the ideal $(x,z,xy-z^2)$ in $k[x,y,z]$, so it suffices to check primality of the latter. But obviously by closure this latter ideal is simply $(x,z)$ which is prime, and so the original ideal is prime.
I find your argument kind of hard to follow and you can see that the above argument is sort of implicit in it.
With the above in mind your second question is not too hard to answer. If $\bar x \in P^2$ then translating back to $k[x,y,z]$ there is some $g\in k[x,y,z]$ such that $$x + (xy -z^2)g \in (x,z)^2 = (x^2,xz,z^2)$$ By closure we can drop the $-z^2g$ to get $$x + xy g \in (x^2,xz,z^2)$$ but the LHS clearly cannot be in that ideal since it has a lone $x$ term.
Finally, for your last question, if $\bar y \in r(P^2) \subseteq r(P)$ then $\bar y^n \in P$ for some $n$ and hence $\bar y \in P$, which would say that $y\in (x,z)$ in $k[x,y,z]$ which is clearly not the case.