This is also directly related to the size of $\sum_{n \leq X} \mu(n)$ as follows :
Assume that the estimate $\sum_{n \leq X} \frac{\mu(n)}{n} = O(\varepsilon(X))$ holds for some non increasing function $\varepsilon$ (with $ \varepsilon(X) = O(\log X ) $ ). From Dirichlet hyperbola principle we have
$$\sum_{n \leq X} \frac{\mu \star \mu(n)}{n} = 2\sum_{n \leq \sqrt{X}} \frac{\mu(n)}{n} \sum_{dn \leq X} \frac{\mu(d)}{d} - \left( \sum_{n \leq \sqrt{X}} \frac{\mu(n)}{n} \right)^2 $$ and thus
$$\sum_{n \leq X} \frac{\mu \star \mu(n)}{n} = O \left( \varepsilon(\sqrt{X}) \log X +\varepsilon(\sqrt{X})^2 \right) = O \left( \varepsilon(\sqrt{X}) \log X \right) $$
Now set $g = \mu \tau \star 1 \star 1$. As noted above by Matt Young and GH, the series $\sum_{n \geq 1} g(n) n^{- \sigma}$ converges absolutely for $\sigma > \frac{1}{2}$. Observe that
$$\sum_{n \leq X} \frac{\mu(n) \tau (n)}{n} = \sum_{n \leq X} \frac{g(n)}{n} \sum_{dn \leq X} \frac{\mu \star \mu(d)}{d} = O \left( \sum_{n \leq X} \frac{|g(n)|}{n} \varepsilon(\sqrt{\frac{X}{n}}) \log \frac{X}{n}X \right) $$$$\sum_{n \leq X} \frac{\mu(n) \tau (n)}{n} = \sum_{n \leq X} \frac{g(n)}{n} \sum_{dn \leq X} \frac{\mu \star \mu(d)}{d} = O \left( \sum_{n \leq X} \frac{|g(n)|}{n} \varepsilon(\sqrt{\frac{X}{n}}) \log \frac{X}{n} \right) $$
Since $\sum_{n \geq X^{\frac{1}{4}}} |g(n)| n^{- 1} = O \left( X^{-\frac{1}{16}} \sum_{n} g(n) n^{- \frac{3}{4}} \right) = O \left( X^{-\frac{1}{16}} \right) $$\sum_{n \geq X^{\frac{1}{2}}} |g(n)| n^{- 1} = O \left( X^{-\frac{1}{6}} \sum_{n} |g(n)| n^{- \frac{2}{3}} \right) = O \left( X^{-\frac{1}{6}} \right) $, we get
$$\sum_{n \leq X} \frac{\mu(n) \tau (n)}{n} = O \left( \varepsilon(X^{\frac{1}{4}}) \log X + X^{-\frac{1}{16}} \right) $$$$\sum_{n \leq X} \frac{\mu(n) \tau (n)}{n} = O \left( \varepsilon(X^{\frac{1}{4}}) \log X + X^{-\frac{1}{7}} \right) $$
For example, an estimate $\sum_{n \leq X} \frac{\mu(n)}{n} = o(\frac{1}{\log X})$ ensures that $\sum_{n \leq X} \frac{\mu(n) \tau (n)}{n} = o \left( 1 \right) $