Let $\mathbb{D}$ denote the open unit disc. In general, an analytic function $f:\mathbb{D}\to\mathbb{C}$ is allowed to have countably many zeros in $\mathbb{D}$. As Friedrich has pointed out,
$$
\sin\left(\frac{1}{1+z}\right)
$$
is an example of a function that is analytic on $\mathbb{D}$ and has infinitely zeros inside $\mathbb{D}$.
However, if we assume that $f$ is continuous on $\mathbb{D}$, and also that $|f(z)| = 1$ for $|z|=1$, then the story changes. Suppose $f$ has countably many zeros $z_n$ in $\mathbb{D}$. Then by compactness, the set $\{z_n\}$ has a limit point in $\overline{\mathbb{D}}$.
The zeros cannot have a limit point on boundary of the unit disc, since if $z_{n_k}\to z_\infty\in\partial\mathbb{D}$ then $f(z_{n_k})\to f(z_\infty)$ by continuity, but $|f(z_{n_k})| = 0$ and $|f(z_\infty)| = 1$, contradiction.
So the limit point in $\mathbb{\overline{\mathbb{D}}}$ must lie inside $\mathbb{D}$. But then $f$ has a sequence of zeros converging inside its domain of definition, and since $f$ is analytic it follows that $f \equiv 0$. This is a contradiction if $f$ is assumed nontrivial.
Therefore it follows that if $f$ is nontrivial, then $f$ can only have finitely many zeros inside $\mathbb{D}$. At this point one can express $f$ as a product of finitely many Blaschke factors using a consequence of the Schwarz lemma.
I'm not sure about that. $f(z)-g(z)=z^7+4z^5-12z^3+z-1$. Then $\lvert f(z)-g(z)\rvert\ne1$ for some $z$ along the unit circle. Even if you wanted $f(z)+g(z)=z^7-z+1$, choosing $z=i$, we get $i^7-i+1=-2i+1$ which has modulus greater than $1$. So, this doesn't work either.
Rather, try comparing the coefficients. We can see that the largest coefficient is $6$, so along the unit circle the term $6z^3$ probably dominates the others. Take $f(z)=6z^3$ and $h(z)=z^7-2z^5-z+1$. Then
$$ \lvert h(z)\rvert\le \lvert z\rvert^7+2\lvert z\rvert^5+\lvert z\rvert +1\le 5,$$
$$\lvert f(z)\rvert= \lvert 6z^3\rvert=6.$$
So, $g(z)$ has as many zeros on the unit disk as $f(z)=6z^3$. $f(z)$ has a zero of multiplicity $3$ at $z=0$, so we conclude that $g$ has $3$ zeros in the unit disk.
Best Answer
If $a=0$ then obviously $z=0$ has multiplicity $n$.
For $a\neq 0$, a root has multiplicity >1 iff $z^n-ae^{-z-1}$ and $nz^{n-1}+ae^{-z-1}$ would both vanish (and for multiplicity 3 or above, $n(n-1)z^{n-2}-ae^{-z-1}$ would also vanish, etc.), but that means $z^n+nz^{n-1}=0$ (adding the two equations). Clearly $z=0$ can't be a root, and the only other possibility is $z=-n$ which lies outside the disc.