Proof

By assumptions, $\alpha,\beta$ are arithmetic functions with $\beta(n)=\sum_{d\mid n}\alpha(d).$
For any divisor $d\mid n$ we have, therefore, $$\beta\left(\frac nd\right)=\sum_{b\mid \frac nd}\alpha(b).$$
Multiplying both sides of the equation by the Möbius function of $\mu(d)$ results in $$\mu(d)\beta\left(\frac nd\right)=\sum_{b\mid \frac nd}\mu(d)\alpha(b).$$
The sum over all divisors $d\mid n$ results in $$\sum_{d\mid n}\mu(d)\beta\left(\frac nd\right)=\sum_{d\mid n}\sum_{b\mid \frac nd}\mu(d)\alpha(b)=\sum_{b\mid n}\sum_{d\mid \frac nb}\mu(d)\alpha(b).$$
In the last step, we could exchange the indices $d$ and $b$ because as the index $b$ runs through all divisors $b\mid \frac nd$ so does $d$ run through all complementary divisors $d\mid \frac nb.$
Now, the term $\alpha(b)$ does not depend on the index $d$ of the inner sum and we can use the distributivity law to get $$\sum_{b\mid n}\sum_{d\mid \frac nb}\mu(d)\alpha(b)=\sum_{b\mid n}\alpha(b)\sum_{d\mid \frac nb}\mu(d).$$
The last sum can be significantly simplified using the sum of Möbius function over divisors and we get $$\sum_{b\mid n}\alpha(b)\sum_{d\mid \frac nb}\mu(d)=\sum_{b\mid n}\alpha(b)[b=n]=\alpha(n).$$
We have shown $$\alpha(n)=\sum_{d\mid n}\mu(d)\beta\left(\frac nd\right).$$
∎

Thank you to the contributors under CC BY-SA 4.0!

Scheid Harald: "Zahlentheorie", Spektrum Akademischer Verlag, 2003, 3rd Edition
Landau, Edmund: "Vorlesungen über Zahlentheorie, Aus der Elementaren Zahlentheorie", S. Hirzel, Leipzig, 1927