**Edit 1:** Looking back at that note, which I wrote $6$ years ago, it's unclear and overly technical. Since $\pi(x)=\frac{x}{\log x}+O\left(\frac{x}{(\log x)^2}\right),$ we have that $$\sum_{p\leq (\log x)^2}\pi\left(\frac{x}{p^3}\right)\sim \frac{x}{\log x}\sum_{p\leq (\log x)^2}\frac{1}{p^3}+O\left(\frac{x\log \log (x)}{(\log x)^2}\right).$$ Now, $$\sum_{p\leq (\log x)^2}\frac{1}{p^3}=P(3)+O\left(\frac{1}{(\log x)^4}\right),$$ so all that remains is to bound the sum $$\sum_{(\log x)^2\leq p\leq x^{1/3}}\pi\left(\frac{x}{p^3}\right)$$ from above, which can be done in a straightforward way with only the bound $\pi(x)=O(x/\log x)$, and we conclude that $$\sum_{qp^3\leq x} 1 \sim P(3)\frac{x}{\log x}.$$

**Edit 2:** Here's a different convoluted way to do it: Sequence of numbers with prime factorization $pq^2$

This can be proven using partial summation. See this note I wrote for the full details. The sum $$\sum_{p} \pi\left(\frac{x}{p^3}\right)=\sum_{p\leq x} \sum_{q\leq \frac{x}{p^3}}1=\sum_{qp^3\leq x} 1$$ counts the number of integers $n\leq x$ such that $n=qp^3$ where $q$ and $p$ are both prime (and not-necessarily distinct). Let's call this $\sigma_{(1,3)}(x)$

You are correct that $$\sigma_{(1,3)}:=\sum_{p} \pi\left(\frac{x}{p^3}\right)\sim P(3)\frac{x}{\log x}.$$ More generally, consider the number of integers of the form $n=p_1^{\alpha_1}\cdot p_2^{\alpha_2}\cdots p_k^{\alpha_k}$ where $1=\alpha_1\leq \alpha_2\leq \dots\leq \alpha_k$ and $1=\alpha_1=\cdots=\alpha_r<\alpha_{r+1}$. Call this $\sigma_{(\alpha_1,\dots,\alpha_r)}$. Then

**Theorem:** We have that $$\sigma_{(\alpha_1,\dots,\alpha_r)}\sim \frac{x(\log \log x)^{r-1}}{\log x}\prod_{i=r+1}^k P\left(\alpha_i\right).$$

This all appears in the note I linked to above.

For example, this means that the number of integers of the form $n=pqr^2s^3$ less than $x$, where $p,q,r,s$ are prime, is asymptotic to $$\frac{x\log \log x}{\log x} P(2)P(3).$$