How to invert this relationship (Möbius function involved)? Background
Let,
$$ S(x,a) = \sum_{r=1}^\infty \sum_{d|r} \frac{x^r}{a^d d!}$$
Another representation of this series for all derivatives around $0$ is given by:
$$ S(x,a) = \sum_{r=1}^\infty (e^{\frac{x^r}{a}} -1 ) $$
where $|x| < 1$ and $a \neq 0$
$$ \frac{1}{2 \pi i} \oint \frac{f(a)}{a}  \lim_{x \to 0 } \frac{\partial^r S}{\partial x^{r}} da = \delta_{1,r}$$
where $r$ is a positive non-zero integer and $\delta_{1,r}$ is only $1$ when $r = 1$
With, $f$ being analytic then we claim:
$$f^{'n}(0) =  \mu(n)$$
where $\mu$ is the Möbius function.
Question
Are there any simplifications which can be done to:
$$ \frac{1}{2 \pi i} \oint \frac{f(a)}{a} \lim_{x \to 0 } \frac{\partial^r S}{\partial x^{r}} da = \delta_{1,r}$$
To possibly invert it?
The below are examples of the claim:
Example $1$
Consider, the following equation:
$$ S(x,a) = \sum_{r=1}^\infty (e^{\frac{x^r}{a}} -1 )$$
Differentiating both sides with respect to $x$:
$$ S'(x,a) = \frac{1}{a}e^{\frac{x}{a}} + \frac{2x}{a}e^{\frac{x^2}{a}} + \dots$$
Multiplying by $\frac{f(a)}{a}$ and taking $x \to 0$:
$$ \frac{f(a)}{a} \lim_{x \to 0}S'(x,a) = \frac{f(a)}{a^2}$$
Taking the contour integral with respect to $a$:
$$\oint \frac{f(a)}{a^2} da = f'(0) = 1$$
Example $2$
Consider, the following equation:
$$ S(x,a) = \sum_{r=1}^\infty (e^{\frac{x^r}{a}} -1 ) $$
Differentiating both sides with respect to $x$ twice:
$$ S''(x,a) = \frac{1}{a^2}e^{\frac{x}{a}} + \frac{2}{a}e^{\frac{x^2}{a}} + \dots$$
Multiplying by $\frac{f(a)}{a}$ and taking $x \to 0$:
$$ \frac{f(a)}{a} \lim_{x \to 0} S''(x,a) = \frac{f(a)}{a^3} + 2 \frac{f(a)}{a^2}$$
Taking the contour integral with respect to $a$:
$$\oint (\frac{f(a)}{a^3}  + 2 \frac{f(a)}{a^2}) da = 2! f''(0) + 2f'(0) \implies f''(0) = -1$$
(Using example $1$ in the last step)
 A: I thought about this for a bit and actually I arrive at a different result. So probably I made a mistake.

Let me change the variables and indices a bit. For $a ∈ ℂ^×$ and $z$ ranging around zero let
$$S(z;a) = \sum_{n=1}^∞ \sum_{d \mid n} \frac{z^n}{a^d d!} = \sum_{ν=1}^∞ (\mathrm e^{\frac {z^ν} a} - 1).$$
Then for any $a ∈ ℂ^×$, $S(z; a)$ is analytical in $z$ around zero and furthermore for $n ≥ 1$
$$S_{n,0}(a) := \lim_{z → 0} \frac {∂^n} {∂z^n} S(z; a) = \sum_{d \mid n} \frac {n!} {a^d d!}.$$
By Cauchy’s integral formula we have for any analytical function $f$ around zero
$$f^{(d)}(0) = \frac {d!} {2π \mathrm i} \oint \frac{f(a)} {a^{d+1}} \mathrm d a,$$
and so
$$\frac 1 {2π \mathrm i} \oint \frac{f(a)} a S_{n,0}(a) \mathrm d a = \sum_{d \mid n}  \frac {n!} {d!}  \frac 1 {2π \mathrm i}\oint \frac {f(a)} {a^{d+1}} \mathrm d a= \sum_{d \mid n} \frac {n!} {d!^2} f^{(d)} (0).$$
Now if for every $n ≥ 1$
$$\frac 1 {2π \mathrm i} \oint \frac{f(a)} a S_{n,0}(a) \mathrm d a = ε(n)$$
with $ε(n) = δ_{n,1}$, then by canceling $n!$, we also get
$$\sum_{d \mid n} \frac{f^{(d)}(0)} {d!^2} = ε(n),$$
and since also $\sum_{d \mid n} μ(d) = ε(n)$, by a simple induction on $n$ we conclude $μ(n) = \frac{f^{(n)}(0)}{n!^2}$.

