We first observe that as $\frac{\mu(p^k)}{\sigma(p^k)}$ is equal to $1$ if $k = 0$, $-\frac{1}{p + 1}$ if $k = 1$, and $0$ otherwise, its Dirichlet series is
\[\sum_{n = 1}^{\infty} \frac{\mu(n)}{\sigma(n) n^s} = \prod_p \left(1 - \frac{1}{(p + 1) p^s}\right) = \frac{R(s)}{\zeta(s + 1)},\]
where
\[R(s) = \prod_p \left(1 + \frac{1}{(p + 1)(p^{s + 1} - 1)}\right).\]
Note that for $\sigma > -1$, the terms in the Euler product for $R(\sigma)$ are of the form $1 + a_p(\sigma)$ with $a_p(\sigma) = \frac{1}{(p + 1)(p^{\sigma + 1} - 1)}$. Since $\sum_p a_p(\sigma)$ converges for $\sigma > -1$, it follows that $R(s)$ is absolutely convergent for $\Re(s) > -1$ and defines a holomorphic function in that region. Moreover, $R(0) = \zeta(2) = \frac{\pi^2}{6}$.

Let $\Theta$ denote the supremum of the real parts of the zeroes of $\zeta(s)$. Suppose in order to obtain a contradiction that
\[\sum_{n \leq x} \frac{\mu(n)}{\sigma(n)} \log \frac{x}{n} - \frac{\pi^2}{6} < x^{-1 + \Theta - \varepsilon}\]
for all $x > x_{\varepsilon}$. Then Landau's lemma implies that if $\sigma_c$ is the infimum of $\sigma \in \mathbb{R}$ for which
\[\int_{1}^{\infty} \left(x^{-1 + \Theta - \varepsilon} - \sum_{n \leq x} \frac{\mu(n)}{\sigma(n)} \log \frac{x}{n} + \frac{\pi^2}{6}\right) x^{-\sigma} \, \frac{dx}{x}\]
is convergent, then
\[\int_{1}^{\infty} \left(x^{-1 + \Theta - \varepsilon} - \sum_{n \leq x} \frac{\mu(n)}{\sigma(n)} \log \frac{x}{n} + \frac{\pi^2}{6}\right) x^{-s} \, \frac{dx}{x}\]
is holomorphic in the right half-plane $\Re(s) > \sigma_c$ but not at the point $\sigma_c \in \mathbb{R}$. On the other hand, this integral is equal to
\[\frac{1}{s + 1 - \Theta + \varepsilon} - \frac{R(s)}{s^2 \zeta(s + 1)} + \frac{\zeta(2)}{s}\]
for $\Re(s) > 0$ and hence for $\Re(s) > \sigma_c$ by analytic continuation. However, this expression has a pole at $s = -1 + \Theta - \varepsilon$ and no other poles on the real line segment $\sigma > -1 + \Theta - \varepsilon$, yet by the definition of $\Theta$, there are poles in the strip $-1 + \Theta - \varepsilon < \Re(s) \leq -1 + \Theta$. Thus a contradiction is obtained, and so it follows that
\[\sum_{n \leq x} \frac{\mu(n)}{\sigma(n)} \log \frac{x}{n} - \frac{\pi^2}{6} = \Omega_+\left(x^{-1 + \Theta - \varepsilon}\right).\]
The same method shows that
\[\sum_{n \leq x} \frac{\mu(n)}{\sigma(n)} \log \frac{x}{n} - \frac{\pi^2}{6} = \Omega_-\left(x^{-1 + \Theta - \varepsilon}\right),\]
so that
\[\sum_{n \leq x} \frac{\mu(n)}{\sigma(n)} \log \frac{x}{n} - \frac{\pi^2}{6}\]
changes sign infinitely often. With a little extra work, we can show that
\[\sum_{n \leq x} \frac{\mu(n)}{\sigma(n)} \log \frac{x}{n} - \frac{\pi^2}{6} = \Omega_{\pm}\left(x^{-1/2}\right).\]

All of this raises the question:

Why does there seem to be a bias?

In other problems like disproving Pólya's conjecture, the bias stems from a term of the same order as the oscillatory terms coming from the zeroes of $\zeta(s)$. But this is not the case in this setting; as Greg mentions in the comments, if we additionally assume the Riemann hypothesis and the Linear Independence hypothesis, then we can show (modulo some details about the growth of $R(s)$ on the line $\Re(s) = -1/2$) that the set
\[\left\{x \in [1,\infty) \colon \sum_{n \leq x} \frac{\mu(n)}{\sigma(n)} \log \frac{x}{n} - \frac{\pi^2}{6} < 0\right\}\]
has a limiting logarithmic density that is equal to $1/2$.

So where does the bias come from? I believe it's due to a lower order term. As $s$ tends to $-1$, $1/\zeta(s+1)$ tends to $-2$, whereas $R(s)$ is positive and blows up as $s \to -1$. In fact, it blows up in two ways: one way from the prime $p = 2$, and the other from the product over the remaining primes. This suggests that $R(s) \sim C/(s + 1)^2$ as $s \to -1$ for some positive constant $C > 0$. So this will give a lower order term of size $-\frac{C' \log x}{x}$ in the asymptotic expansion of $\sum_{n \leq x} \frac{\mu(n)}{\sigma(n)} \log \frac{x}{n} - \frac{\pi^2}{6}$, where $C' > 0$ is some positive constant.

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