When yesterday I was interested in do a little study about the arithmetic function $$f(n)=\left( 1+|\mu(n)| \right)^{M(n)},$$ defined for integers $n\geq 1$, which $\mu(n)$ is the Möbius function and $M(x)=\sum_{k\leq x}\mu(k)$, thus $M(n)$ is the Mertens function, I have doubts about if my computations were interesting and if it is possible improve my study, but the Question (see below) is the main purpose for this post.
Combining the fact that $\log \left( 1+|\mu(n)| \right)$ is positive and bounded by $\log 2$ with the Prime Number Theorem I can prove that $$\log f(n)=o(n).$$
I know from Wikipedia's Page for square-free integers the (obvious) bound $$\mathcal{Q}(n)=\frac{6}{\pi^2}n+O \left( \sqrt{n} \right) ,$$ where $\mathcal{Q}(x)$ is the number of square-free integers between 1 and $x$. I say this fact since I will accept it if you use/need it.
Question. Can you get the asymptotyc behaviour (I say for example a big oh result, that is an asymptotic identity with error term denoted with a big oh notation) for $$\sum_{\substack{n\leq x\\ n\text{ square-free}}}2^{M(n)}$$ as $x\to\infty$? Thus I am asking about the average order for $f(n)$. Thanks in advance.
With this exercise my purpose is learn about somethings related with sieves, if you can use it in your answer.
I don't know if such kind of questions were in the literature, perhaps the definition about $f(n)$ hasn't interesting properties.
Appendix: (Is only a comment): And this was my attempt to combine the definition of previous function $f(n)$ with the Riemann hypothesis:
An additional doubt is if the abscisse $\Re s>\frac{3}{2}$ is the best that I can state on assumption of the Riemann hypothesis when I want to state $$\sum_{n=2}^{\infty}\frac{M(n)}{n^s}-\sum_{n=2}^{\infty}\frac{\log \left( f(n-1) \right) }{ \left( \log \left( 1+|\mu(n-1)| \right) \right)n^s }=\frac{1}{\zeta(s)}-1.$$