I will assume we can write
$$f(z) = \frac{c}{z_0-z} + \sum_{n=0}^{\infty} b_n z^n$$
for some value of $c \ne 0$, and $\lim_{n \to \infty} b_n = 0$. Then
$$f(z) = \sum_{n=0}^{\infty} a_n z^n$$
where
$$a_n = b_n + \frac{c}{z_0^{n+1}}$$
Then
$$\begin{align}\lim_{n \to \infty} \frac{a_n}{a_{n+1}} &= \lim_{n \to \infty} \frac{\displaystyle b_n + \frac{c}{z_0^{n+1}}}{\displaystyle b_{n+1}+ \frac{c}{z_0^{n+2}}}\\ &= \lim_{n \to \infty} \frac{\displaystyle \frac{c}{z_0^{n+1}}}{\displaystyle \frac{c}{z_0^{n+2}}}\\ &= z_0\end{align}$$
as was to be shown. Note that the second step above is valid because $z_0$ is on the unit circle.
For a nonsimple pole, we may write
$$f(z) = \frac{c}{(z_0-z)^m} + \sum_{n=0}^{\infty} b_n z^n$$
for $m \in \mathbb{N}$. It might be known that
$$(1-w)^{-m} = \sum_{n=0}^{\infty} \binom{m-1+n}{m-1} w^n$$
Then
$$\lim_{n \to \infty} \frac{a_n}{a_{n+1}} = \lim_{n \to \infty} \frac{n+1}{n+m} z_0$$
EDIT
@TCL observed that we can simply require that $b_n z_0^n$ goes to zero as $n \to \infty$. Then for a simple pole
$$\lim_{n \to \infty} \frac{a_n}{a_{n+1}} = \lim_{n \to \infty} \frac{\displaystyle b_n z_0^n + \frac{c}{z_0}}{\displaystyle b_{n+1} z_0^n + \frac{c}{z_0^2}}$$
which you can see goes to $z_0$.
For part a), we can simply take $f_n(z) = \frac{z}{n}$, then $f_n(\mathbb{D}) = D_{1/n}(0)$, and since $\frac{1}{n}\to 0$, the result follows. We have $f_n'(0) = \frac{1}{n}\to 0$, which makes it easy.
Part b), which demands we restrict our attention to functions with $f'(0) = 1$, is more interesting. It is easy to find a(n entire, if we desire so) holomorphic $f$ with $f(0) = 0$, $f'(0) = 1$ and $-1\notin f(\mathbb{D})$; $f(z) = \sin z$ will do, or even $f(z) = z$. Then $1+f(z)$ has no zeros in $\mathbb{D}$, and therefore
$$g_n(z) = (1+f(z))^n - 1$$
satisfies $g_n(0) = 0$ and does not attain the value $-1$ on $\mathbb{D}$. But $g_n'(0) = n(1+f(0))^{n-1}\cdot f'(0) = n\cdot 1^{n-1}\cdot 1 = n$, so
$$f_n(z) = \frac{1}{n}g_n(z) = \frac{(1+f(z))^n-1}{n}$$
is holomorphic where $f$ is, satisfies the two conditions $f_n(0) = 0$ and $f_n'(0) = 1$, and does not attain the value $-\frac{1}{n}$ on $\mathbb{D}$.
Best Answer
Note that $a=1$. Then rewrite your equation as
$$z=e^z P(z)$$
where $P(z)$ is a polynomial of degree $n$. For $|z|$ large enough, $P$ grows of order $|z|^n$. Therefore this equation implies upon taking absolut values that $e^{\mathrm{Re} z}$ decreases like $|z|^{1-n}$ as $|z|\rightarrow\infty$. This is clearly a contradiction.
Therefore you cannot apply Hadamard's theorem, i.e. $f$ has infinitely many zeros.