The claim
It follows that, $\forall \Pi,\Pi' \in NP$, if $\Pi'$ is strongly NP-complete, and $\Pi' \leq_{pp} \Pi$, then $\Pi$ is strongly NP-complete.
is wrong (at least for how "pseudo-polynomically reducible" is defined here).
If $\Pi'$ is strongly $NP$-complete, then ``pseudo-polynomially reducible'' is equivalent to polynomially reducible.
Recall that a problem is strongly $NP$-complete if it is $NP$-complete even if we assume all the integers that appear in the input are polynomial in the length of the input.
Recall that an algorithm runs in pseudopolynomial time if it runs in polynomial time in the integers that appear in the input.
So, a pseudopolynomial reduction must run in time polynomial in the integers that appear in the input. Which, for a strongly $NP$-complete problem, can be assumed to be polynomial in the length of the input. Which means the reduction must run in time polynomial in the length of input.
So, if $\Pi'$ is strongly $NP$-complete, $\Pi' \leq_{pp} \Pi$ is equivalent to $\Pi' \leq_p \Pi$. Clearly, the existence of such a reduction does not imply that $\Pi$ is strongly $NP$-hard.
The only place in the literature that I can find any mention of this notion of ``pseudopolynomial reduction'' is in the appendix an OR & Management Science book. I cannot find any other literature mentioning this notion; one would expect to be able to find some mention of it in complexity theory circles.
We should use a different definition for $\leq_{pp}$. A pseudopolynomial reduction should be defined as a reduction, running in pseudpolynomial time, that in addition does not increase integer values in the input more than polynomially. Then the claim would hold true.