What is unclear to me is how you go from $(fg)^{-1}$ to $f^{-1}g^{-1}$. Given that we are dealing with Dirichlet inverses, this is not obvious to me. In fact, it's not true. For example, if $f=g=u$, then $(fg)^{-1} = u^{-1} = \mu$, but $f^{-1}g^{-1} = \mu\mu\neq \mu$.
So that is where the error in your reasoning comes, I think.
Remember that since ${}^{-1}$ represents the Dirichlet inverse, which is the inverse of convolution, you know that $(f*g)^{-1} = g^{-1}*f^{-1} = f^{-1}*g^{-1},$ but you are trying to apply this to the pointwise product, which is incorrect.
As to proving the desired identity, if $h$ is an arithmetic function with $h(1)\neq 0$, then $h^{-1}$ is given by: $ h^{-1}(1) = \frac{1}{h(1)},\qquad h^{-1}(n) = \frac{-1}{h(1)}\sum_{{d|n},\,{d\lt n}}h\left(\frac{n}{d}\right)h^{-1}(d),\quad n\gt 1.$
If $h$ is completely multiplicative then you have $f^{-1}(n)=\mu(n)f(n)$ for all $n\geq 1$.
So we verify that the identity holds for $f$ completely multiplicative and $g$ an arithmetic function with $g(1)\neq 0$.
At $n=1$, we have $(f\cdot g)^{-1}(1) = \frac{1}{f(1)g(1)} = \frac{1}{g(1)}$ (since $f(1)=1$ must hold); and $(f\cdot g^{-1})(1) = f(1)g^{-1}(1) = \frac{1}{g(1)}.$
If $n\gt 1$ and the result hold for all integers smaller than $n$, then we have: $\begin{align*} (f\cdot g)^{-1}(n) &= \frac{-1}{(f\cdot g)(1)} \sum_{d|n, d\lt n}(f\cdot g)\left(\frac{n}{d}\right)(f\cdot g)^{-1}(d)\\ &= \frac{-1}{g(1)} \sum_{d|n\,d\lt n} f\left(\frac{n}{d}\right)g\left(\frac{n}{d}\right) f(d)g^{-1}(d)\\ &= \frac{-1}{g(1)}\sum_{d|n\,d\lt n}\frac{f(n)}{f(d)}g\left(\frac{n}{d}\right)f(d)g^{-1}(d)\\ &= \frac{-f(n)}{g(1)}\sum_{d|n\,d\lt n}g\left(\frac {n}{d}\right)g^{-1}(d)\\ &= f(n)\left(-\frac{1}{g(1)}\sum_{d|n\,d\lt n}g\left(\frac{n}{d}\right)g^{-1}(d)\right)\\ &= f(n)g^{-1}(n)\\ &= (f\cdot g^{-1})(n). \end{align*}$ We used the fact that $f$ is completely multiplicative to get that $f\left(\frac{n}{d}\right) = \frac{f(n)}{f(d)}$. Equivalently, you can note that $f\left(\frac{n}{d}\right)f(d) = f(n)$.
This proves the first part of the exercise.
The second part of the exercise asks you to show that if $f$ is multiplicative and $(f\cdot \mu^{-1})^{-1} = f\cdot \mu$ holds, then $f$ is completely multiplicative.
Since $f$ is completely multiplicative if and only if $f^{-1} = \mu\cdot f$, you are being asked to show that $f\cdot \mu^{-1} = f$. That is, you need to show that $\mu^{-1}(n) = 1 = u(n)$.
But we know this, since $\sum_{d|n}\mu(d) = I(n)$ (the identity of the convolution), so $u$ and $\mu$ are inverses of each other.