Regarding your context:
Let $S$ be the set of zeroes of $g$.
By the inequality, the set of zeroes of $f$ is also $S$.
The domain of an entire function is necessarily $\mathbb C$ by definition. Therefore, barring any sort of ‘extensions’, the largest possible domain of $\frac fg$ is $\mathbb C\setminus S$, due to the fact that $\frac fg=\frac 00$ on $S$ and $\frac 00$ is not well-defined. Thus, $\frac fg$ cannot be entire.
The statement you want to prove is ‘$f$ and $g$ are multiple of each other.’ Mathematically, this can be restated as $f=cg$ for some universal, non-zero constant $c$.
This statement is trivially true on $S$, what remains is proving it on $\mathbb C\setminus S$.
You may proceed like this:
Let $S$ be the set of zeroes of $g$.
By the inequality,
$$\left\vert\frac fg\right\vert \le 1\text{ for }\mathbb C\setminus S$$
Let $h=\frac fg$. Since the zeroes of $g$ is isolated, there exist a neighbourhood $N$ of every element of $S$, such that $N\in\mathbb C\setminus S$ and thus $|h|\le 1$ holds on $N$.
By Riemann’s removable singularity theorem, $h$ can be extended to an entire $H$.
Then, by Liouville theorem $H=c$ on $\mathbb C$.
Recall that $H=h$ on $\mathbb C\setminus S$. Hence $h=c$ on $\mathbb C\setminus S$.
Therefore, you can conclude $f=cg$ on $\mathbb C\setminus S$.
A few final words: Your first question regarding $f$ cannot be answered because you did not specify how $f$ is defined on $A$.
Whenever you ask whether a function $f$ is entire, always think about where did you define it. A function is always defined along with a domain, and $f$ can be entire only if its domain is $\mathbb C$.
If you define $\sin z :[0,1]$, it can never be entire. If you have $f$ holomorphic on $\mathbb C\setminus A$, before you ask whether it is entire, ask yourself how $f$ is defined on $A$. If for $a\in A$, $f(a)$ does not return a complex number but a set, or a function, or $\text{Donald Trump}$, then there is no point to discuss about being entire or not.
It turns out that it is the same case in your context: without any extensions, $\frac fg$ cannot be defined on $S$ because we don’t know how to define $\frac 00$. Discussion on entireness immediately ends. Of course, if you define $\frac fg$ on $S$ by its continuous extension, then by Riemann’s removable singularity theorem continuous extension is same as holomorphic extension, hence $\frac fg$ is holomorphic on $S$ too.
Best Answer
Consider $\frac{f(z)}{z^n}$. Since $f$ was entire, we just need to show that the singularity at $0$ is removable. Because of Reimann's theorem(read here https://en.wikipedia.org/wiki/Removable_singularity#Riemann.27s_theorem) this is equivalent to showing that
$$\lim_ {z \to 0}z\frac{f(z)}{z^n} = \lim_ {z \to 0}\frac{f(z)}{z^{n-1}} =0 $$ But this follows from your information that $|f(z)| \leq |z|^n$.
Rest of your proof was correct. As far as $n$ is concerned, it will be equal to your earlier $n$.(Prove this !)