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For a set $C$, write $2\cdot C=\{2c:c\in C\}$. This is a dilate, whereas $C+C$ allows two independently chosen elements.

A minimal-growth lemma. Let $A$ and $V$ be nonempty finite sets in an abelian group. Choose a nonempty $X\subseteq V$ minimizing $\kappa=|X+A|/|X|$ over nonempty subsets of $V$. Then for every finite set $C$,

$$|X+A+C|\le\kappa|X+C|.$$

Here is the additive proof of Petridis's Proposition 2.1. Order $C=\{c_1,\ldots,c_t\}$ and let $X_i=\{x\in X:x+c_i\notin X+\{c_1,\ldots,c_{i-1}\}\}$. The new elements at step $i$ in $X+C$ are exactly $X_i+c_i$. In $X+A+C$, all elements of $(X\setminus X_i)+A+c_i$ have already appeared. Thus the number of new elements there is at most

$$|X+A|-|(X\setminus X_i)+A| \le\kappa|X|-\kappa|X\setminus X_i| =\kappa|X_i|.$$

The inequality uses minimality, with an empty complement contributing zero. Summing over $i$ proves the lemma. Taking $C=A$ gives $|X+A+A|\le\kappa^2|X|$.

A third-projection inequality. Suppose $A+B$ has unique representations, and put $t=|A+2\cdot B|$. Apply the lemma with $V=2\cdot B$ and the summand set $A$. Then $\kappa\le t/s$. For its nonempty $X\subseteq2\cdot B$, the sums $2a+x$ are all distinct as $(a,x)$ varies: writing $x=2b$, equality reduces to $a+b=a'+b'$. Hence

$$r|X|=|2\cdot A+X|\le|A+A+X| \le\kappa^2|X|\le(t/s)^2|X|.$$

Canceling $|X|$ gives $t\ge s\sqrt r$. Interchanging $A$ and $B$ proves $|2\cdot A+B|\ge r\sqrt s$. These inequalities hold for any nonempty subrectangle as well.

The multiple layers. In the stated prime grid, the multiples of $p_{ij}$ are

$$M_{ij}(t)=z-a_i+t(q+a_i+b_j).$$

The row and column locations are $R_i=M_{ij}(0)=z-a_i$ and $C_j=M_{ij}(1)=z+q+b_j$. Write

$$E_{ij}=M_{ij}(-1)=z-q-2a_i-b_j,\qquad D_{ij}=M_{ij}(2)=z+2q+a_i+2b_j.$$

The preceding locations, the row locations, the column locations and the following locations lie respectively in

$$\begin{aligned} &[z-q-2a_r-b_s,z-q], &&[z-a_r,z],\\ &[z+q,z+q+b_s], &&[z+2q,z+2q+a_r+2b_s]. \end{aligned}$$

These intervals are pairwise disjoint under $2a_r+3b_s<q$. If $J$ contains the row and column locations and has at least $3P$ integers, then for each cell it contains $E_{ij}$ or $D_{ij}$. Otherwise, since it contains $R_i$, it would lie strictly between $E_{ij}$ and $D_{ij}$, an interval containing only $3p_{ij}-1<3P$ integers.

A median rectangle. Put $i_0=\lceil r/2\rceil$, $j_0=\lceil s/2\rceil$, and let $U$ be the right endpoint of $J$. If $U\ge D_{i_0j_0}$, all $D_{ij}$ with $i\le i_0$, $j\le j_0$ belong to $J$. They lie after the core and at or before $U$. The third-projection inequality on this lower-left rectangle gives at least

$$j_0\sqrt{i_0}\ge\frac{s\sqrt r}{2\sqrt2}$$

distinct following locations. If $U<D_{i_0j_0}$, every $D_{ij}$ with $i\ge i_0$, $j\ge j_0$ is outside $J$, so every corresponding $E_{ij}$ belongs to $J$. The other inequality on this upper-right rectangle gives at least

$$(r-i_0+1)\sqrt{s-j_0+1}\ge\frac{r\sqrt s}{2\sqrt2}$$

distinct preceding locations. In either case these locations are disjoint from the $r+s$ core locations. This proves the stated count.

For the power bound, write $m=\max(r,s)$, $n=\min(r,s)$ and $u=mn$. Then $\min(s\sqrt r,r\sqrt s)=n\sqrt m=u/\sqrt m$. With $c=1/(2\sqrt2)$ the count is at least $m+cu/\sqrt m$. The minimum of this expression over positive real $m$ is $3(cu/2)^{2/3}=3u^{2/3}/2^{5/3}$.

Replacing $z$ by $z+v\prod_{i,j}p_{ij}$ for a sufficiently large integer $v$ makes any fixed block positive and preserves its count.