I've verified conjectured relationship (3) below for several thousand values of $x$ and suspect that it's true.
(1) $\quad H(x)=\sum\limits_{n=1}^x\frac{1}{n}\qquad\qquad\qquad\quad$(Harmonic Number Function)
(2) $\quad M(x)=\sum\limits_{n=1}^x\mu(n)\qquad\qquad\qquad$(Mertens Function)
(3) $\quad H(x)\stackrel{?}{=}\sum\limits_{n=1}^x\frac{\sigma_1(n)}{n}\,M\left(\frac{x}{n}\right)\qquad\quad(\sigma_1(n)$ is the Sum-of-Divisors Function$)$
Question: Has conjectured relationship (3) above been proven or disproven (or can it be)?
Yes, this relationship is true and can be proved as follows (FYI $\sigma_1(n)=\sum_{d\mid n}d$ is more commonly known as the sum of divisors function, while the divisor function tends to refer to $\sum_{d\mid n}1$).
$$ \sum_{1\leq n\leq x}\frac{\sigma_1(n)}{n}M(x/n) = \sum_{1\leq n\leq x}\frac{1}{n}\sum_{d\mid n}d\sum_{1\leq m\leq x/n}\mu(m). $$
The right-hand side is
$$ = \sum_{1\leq d,m\leq x} d\mu(m)\sum_{\substack{1\leq n\leq x/m\\ d\mid n}} \frac{1}{n} = \sum_{1\leq dmn\leq x} \frac{\mu(m)}{n}= \sum_{1\leq k\leq x}\mu \ast 1\ast f(k)$$
where $f(t)=1/t$ and $\star$ denotes the multiplicative Dirichlet convolution. By Mobius inversion $\mu\star 1$ is the identity function for convolution, and hence this sum is just $\sum_{1\leq k\leq x}f(k)=H(x)$ as you claim.