Here is a purely mathematical version of f''’s answer.
First note that the functions $s(x,y)$ and $p(x,y)$ always take non-negative inputs to positive outputs; hence, so does any operation build up from them. (To make this argument fully rigorous, it’s probably most natural to consider them as total binary operations on the extended non-negative real line $\{ x \in \mathbb{R}\ |\ x \geq 0 \} \cup \{\infty\}$, with $p(x,\infty)$ and $p(\infty,x)$ defined to be $x$; or, even better, on the projective line $\mathbb{P}^1(\mathbb{R}_{>0})$. But I won’t quite be that formal here.)
Now I claim: for any function $f(x,y)$ built up from the operations $s$ and $p$ together with non-negative constants, if $f(0,y) = 0$ for all $y$, then $f(x,y) = 0$ for all $x,y$.
Suppose this failed. Then there would be some minimal counterexample $f(x,y)$ — that is, minimal in its complexity as a formula built from $s$ and $p$, not necessarily minimal in its numerical output values.
It can’t be a constant, so it’s either of the form $s(g(x,y),h(x,y))$ or $p(g(x,y),h(x,y))$, where $g$ and $h$ are simpler expressions, so in particular are not counterexamples to the claim.
First case: $f(x,y) = s(g(x,y),h(x,y))$. So $g(0,y) + h(0,y) = 0$ for all $y$. So, since everything involved is non-negative, $g(0,y) = h(0,y) = 0$, for all $y$; so since $g$ and $h$ were not counterexamples to the claim, $g(x,y) = h(x,y) = 0$ for all $x,y$. So $f(x,y) = 0$ always — contradicting the choice of it as a counterexample!
Second case: $f(x,y) = p(g(x,y),h(x,y))$. (This case gets a bit mathematically involved.) Now $p(a,b) = 0$ exactly if at least one of $a,b$ is 0. So we know that for all $y$, at least one of $g(0,y)$ and $h(0,y)$ is $0$. So either $g(0,y)$ is zero for a dense set of $y$, or $h(0,y)$ is zero on some non-empty open set of $y$’s. But these are rational functions — and if a rational function is 0 either on a dense set or on a non-empty open set, then it must be zero everywhere. So either $g(0,y)$ is always zero, or $h(0,y)$ is. Since $g$ and $h$ aren’t counterexamples, this means that either $g(x,y)$ is always zero, or $h(x,y)$ is. But either of these means that $f(x,y)$ is always zero. Contradiction again!
So no such function is possible.