One can show that the Prime Number Theorem is equivalent to the statement $ A(x):= \sum_{n \leq x} \frac{\mu(n)}{n}=o(1),\qquad \qquad (1)$ i.e. that $A(x) \to 0$ as $x \to \infty$. Given that the identity $\sum_{n=1}^\infty \frac{\mu(n)}{n^s} = \frac{1}{\zeta(s)} \qquad\qquad \qquad (2)$ holds (elementary) for $\Re s >1$, line (1) seems to encode nothing more than the fact that $\zeta(s)$ is non-vanishing on the critical line $\Re s=1$ (equivalent to the PNT). By analogy, I would expect that $\sum_{n=1}^\infty \frac{\mu(n)\log n}{n^s} = \frac{d}{ds} \frac{1}{\zeta(s)}=\frac{-\zeta'(s)}{\zeta(s)^2} \qquad \qquad (3)$ to hold not only for $\Re s >1$, but for $\Re s =1$ (once again, using nothing more than the PNT). If so, then $\sum_{n=1}^\infty \frac{\mu(n) \log n}{n}=1 \qquad \qquad \quad(4)$ using (3) while ignoring the removable singularity at $s=1$. How can I back up this intuition?
I know that (4) holds under the Riemann Hypothesis, but I believe (and would consequently like to show) that the PNT alone should suffice.