EDIT: I believe I've figured it out! Feel free to take a look in case I've made a mistake.
Problem: Let $n$ and $d$ be positive integers and $m, b \in R$, some ring. If $F(n) = \sum_{d|n}f(d)$ and $\sum_{d|n}g(d) = mF(n)+b$, then $g(n) = mf(n) + b \iota(n)$.
Proof:
$$\sum_{d|n}g(d) = mF(n)+b$$ $$\Rightarrow$$ $$g(n) = \sum_{d|n}\mu(d)(mF(\frac{n}{d})+b)=m\sum_{d|n}\mu(d)F(\frac{n}{d})+b\sum_{d|n}\mu(d)$$ But our other assumption is, $F(n) = \sum_{d|n}f(d) \Rightarrow f(n)=\sum_{d|n}\mu(d)F(\frac{n}{d})$. Then, $$m\sum_{d|n}\mu(d)F(\frac{n}{d})+b\sum_{d|n}\mu(d) = mf(n) + b\sum_{d|n}\mu(d)$$ $$\Rightarrow$$ $$g(n) = mf(n) + b\sum_{d|n}\mu(d)$$
But $\iota (n) = 1$ if $n=1$ and $\iota (n) = 0$ if $n>1$, so $\iota(n)$ and $\sum_{d|n}\mu(d)$ have the same property. So we conclude, $$g(n) = mf(n) + b\iota(n)$$
From the comments I will try to put you in the track:
You correctly have tried to use the Mobius inversion formula getting that $$f(n)=\sum _{d|n}\mu (n/d)F(d),$$ so, if you try to compute $\sum _{d|n}\mu (n/d)(mF(d)+b),$ and you show that $\iota (n)=\sum _{d|n}\mu (d),$ Hint: Show it for a power of prime and use multiplicativity, then $$\sum _{d|n}\mu (n/d)(mF(d)+b)=m\sum _{d|n}\mu (n/d)F(d)+\sum _{d|n}\mu (n/d)b=mf(n)+b\iota(n),$$ concluding, again by Mobius, that this is $g(n).$