In this question of some years ago on MO, the presence of a negative bias for the Mertens function was hypothesized.
A key point for such a problem is how the bias is defined. For example, if we focus on the comparison between the percentage of values of $M(n)$ that are positive or negative for $n ≤ x$, as noted in one of the answers this article conditionally proves that the function is unbiased, as positive and negative points have the same logarithmic density. A nice illustration in section 6 of this paper confirms that, although a negative bias seems to be evident up to $x=10^5$, for $10^5 \leq x \leq 10^{16}$ such bias is no longer present.
As a different example, an apparent bias for the Mertens function seems to emerge by simple summation methods, often used to assign values to divergent series. In this other question on MO, standard approaches as Euler or Hölder summations were applied, and the resulting plots suggested that the divergent $M(n)$ series could be evaluated at $-2$. Although the sums were calculated only with the first $128$ values of $M(n)$, the plots seem to show a relatively rapid convergence to this value. This answer noted that, by $\zeta$-regularization, starting from the well known relation
$$\sum_{n=1}^\infty\frac{\mu(n)}{n^s}=\frac1{\zeta(s)}$$
for $\Re(s)>1$ and setting $s=0$, one would evaluate the infinite sum of $\mu(n)$ at $1/\zeta(0)=-2$.
Another example was given in this paper, which noted a negative bias by showing that
$$\lim_{x\to1^+} \sum_{n=1}^\infty \mu(n)\frac{x^n}{1+x^n} = -1$$
Here one could note that the limit of the fraction is $1/2$, somewhat in agreement with the estimate of $-2$ given above to the sum of the Mobius function. Overall, it is interesting to note that different methods to assign a value to a divergent series such as $M(x)=\sum_{n\leq x} \mu(n)$ give similar results.
I tried to investigate this issue further, considering the values of the Mertens function for $n\leq 30000$ and using iterated Cesàro summations. In the figure below, the top graph shows $M(n)$, while the bottom graph shows the corresponding values of successive sums defined as
$$S_1(x)=\frac{1}{x} \sum_{n\leq x} M(n) \,\text{ (orange)}$$ $$S_2(x)= \frac{1}{x} \sum_{n\leq x} S_1(n) \,\text{ (light blue)} $$ $$S_3(x)= \frac{1}{x} \sum_{n\leq x} S_2(n) \,\text{ (black)} $$
Interestingly, the series $S_1(x)$ is still clearly divergent. Albeit less evidently, $S_2(x)$ seems also to be divergent. Regarding $S_3$, the plot might suggest convergence to $-2$, but it is possible that divergence becomes evident for higher values of $x$.
Questions:
Is there a way to show whether $k$ exists such that $S_k(x)$ is convergent?
Irrespective of the convergence, the $S_k(x)$ series oscillate around the value of $-2$, in accordance with the other approaches. How should we interpret this? Maybe the explicit Titchmarsh formula
$$M(x) = -2 + \sum_\rho \frac{x^\rho}{\rho \zeta'(\rho)} + \sum_{k \ge 1} \frac{x^{-2k}}{-2k \zeta'(-2k)}$$
assuming the RH and the simplicity of the non-trivial zeros, could be useful for this purpose.