Legendre's sieve (inclusion–exclusion count)

Every integer $a\in A$ is coprime to $P(z)$ iff it is divisible by none of the primes $p<z$. For each divisor $d\mid P(z)$ (a squarefree product of some subset of these primes), the count of multiples of $d$ in $A=\{1,\ldots,x\}$ is exactly $\lfloor x/d\rfloor$.

By inclusion–exclusion over the events "$p\mid a$" for $p<z$: the count of $a\in A$ divisible by at least one prime in $\mathcal P$ is $\sum_{p<z}\lfloor x/p\rfloor - \sum_{p<q}\lfloor x/(pq)\rfloor + \sum_{p<q<r}\lfloor x/(pqr)\rfloor - \cdots$, alternating with the number of primes multiplied together. Each such alternating sign is exactly the Möbius function $\mu(d)$ of the corresponding squarefree $d\mid P(z)$: $\mu(1)=1$, $\mu(d)=(-1)^k$ for $d$ a product of $k$ distinct primes.

So the count divisible by at least one prime in $\mathcal P$ is $-\sum_{d\mid P(z),\,d>1}\mu(d)\lfloor x/d\rfloor$. Subtracting this from $|A|=\lfloor x\rfloor=x$ (the $d=1$ term, $\mu(1)\lfloor x/1\rfloor=x$) gives the count coprime to $P(z)$: $S(A,\mathcal P,z)=x-\sum_{d\mid P(z),d>1}\mu(d)\lfloor x/d\rfloor=\sum_{d\mid P(z)}\mu(d)\lfloor x/d\rfloor$.

Finally, taking $z=\sqrt x+1$ so that $\mathcal P$ is exactly the primes $\le\sqrt x$: any $a\in[2,x]$ coprime to all of them is either $1$ or a prime $>\sqrt x$ (by the theorem above, since it has no prime factor $\le\sqrt x$), so $S(A,\mathcal P,\sqrt x+1)=1+\big(\pi(x)-\pi(\sqrt x)\big)$, giving the stated identity.