The proof of Lemma A.9 in Opera de Cribro appears to be wrong or ambiguous

Question

When reading the book Opera de Cribro, In the appendix A.5, I can't see how the proof of Lemma A.9 works. (page 499) Here, $g(d)$ is a multiplicative function with $0\le g(p)<1$ and $$\sum_{p\le x} g(p)\log p=\log x + c_g + O((\log x)^{-A}))$$ where $A\ge 0$ and defined two multiplicative functions given by $\Delta(p^\alpha)=\Delta(p)=1-g(p)p$ and $\delta(a)=\frac{\Delta(a)}{a}$ (this definition is used to produce a convergent series latter). We can prove (which the book asserted and omitted the proof, and I have verified these assertions) $$\sum_{p^\alpha>z}\delta(p^\alpha)\ll (\log z)^{-A}$$ and $$\sum_{z<p^\alpha\le y}\Delta(p^\alpha)\ll y(\log z)^{-A}.$$ Then Lemma A.9 asserts that $$\sum_{a\le y} \Delta(a)\ll y(\log y)^{-A}$$ using the Buchstab formula, written on the book $$\sum_{a\le y}\Delta(a)=1+\sum_{1\le m\le y}\Delta(m)\sum_{q_m<p^\alpha\le y/m}\Delta(p^\alpha)$$ where $q_m$ is the largest prime power of $m$, and $r=\omega(m)$, we can easily get $q_m>m^{1/r},m\le y^{r/(r+1)}$ so $$\sum_{q_m<p^\alpha\le y/m}\Delta(p^\alpha)\ll \frac{y}{m}(\log \frac{y}{m})^{-A}\ll \frac{y}{m}(\frac{r+1}{\log y})^{A}\ll \frac{\tau(m)y}{m(\log y)^A}$$ so far everything seems good, and then they used the trivial bound $$\sum_{m\le z} |\delta(m)|\tau(m)\ll\prod_{p<z}(1+2|\delta(p)|+\dots)\ll (\log z)^4$$ and asserts it can produce the desired result in A.9. But this does not seem to work anyway, because if we let $z=y$ and use the bound, we should get $$\sum_{a\le y}\Delta(a)\ll \sum_{m\le y} |\delta(m)| \tau(m) \frac{y}{(\log y)^A}\ll y(\log y)^{4-A}$$ instead, which is a little bit worse than the asserted bound $y(\log y)^{-A}$. Does anyone have an idea how this could be worked around or is it an error? Thanks! (all letter $p$ denotes prime number)