Here is an example from Which sets are removable for holomorphic functions?
The function $f=z+z^{-1}$ maps the punctured unit disk $\mathbb D$ bijectively onto $\mathbb C\setminus [-2,2]$. The inverse $g=f^{-1}$ is holomorphic in $\mathbb C\setminus [-2,2]$ and is bounded by $1$, but has no holomorphic (or even continuous) extension to $\mathbb C$. Indeed, $g(z)$ approaches both $i$ and $-i$ as $z\to 0$.
A compact set $K$ is removable for bounded holomorphic functions if and only if its analytic capacity is zero. A simple sufficient condition was given by Painlevé: if the $1$-dimensional Hausdorff measure of $K$ is zero, then it's removable for bounded holomorphic functions. That is, every bounded holomorphic function on $\mathbb C\setminus K$ extends to a holomorphic function on $\mathbb C$ (which is necessarily constant by Liouville's theorem).
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
Let $\gamma: [0, 2 \pi] \to \Bbb C$ be the circle $\gamma(t) = P + \rho e^{it}$ for some $\rho \in (0, r)$. Cauchy's integral formula holds for every extension $\hat f_n$ of $f_n$, therefore $$ f_n(z) = \hat f_n(z) = \frac{1}{2 \pi i} \int_\gamma \frac{\hat f_n(\zeta) \, d\zeta}{\zeta - z} = \frac{1}{2 \pi i} \int_\gamma \frac{f_n(\zeta) \, d\zeta}{\zeta - z} $$ for $0 < |z - P | < \rho$ and all $n$. And since $f_n \to f$ uniformly on the image of $\gamma$ it follows that $$ f(z) = \frac{1}{2 \pi i} \int_\gamma \frac{f(\zeta) \, d\zeta}{\zeta - z} $$ for $0 < |z - P | < \rho$. The right-hand side is a continuous (even holomorphic) function in $D(P,\rho)$, which means that $f$ can be holomorphically extended at $P$ as well.
Alternatively one can argue as follows: Each $f_n$ and $f$ can be developed into Laurent series at $z=P$: $$ f_n(z) = \sum_{k=-\infty}^\infty a_{k}^{(n)} (z-P)^k \, , \, f(z) = \sum_{k=-\infty}^\infty a_{k} (z-P)^k $$ and the coefficients can be computed with a generalized Cauchy integral formula: $$ a_{k}^{(n)} = \frac{1}{2 \pi i}\int_\gamma \frac{f_n(z) \, dz}{(z-P)^{n+1}} \, , \, a_{k} = \frac{1}{2 \pi i}\int_\gamma \frac{f(z) \, dz}{(z-P)^{n+1}} $$ The locally uniform convergence implies that $$ \lim_{n \to \infty} a_{k}^{(n)} = a_k $$ for all $k \in \Bbb Z$. If each $f_n$ has a removable singularity at $z=P$ then $a_{k}^{(n)} = 0$ for all $n$ and all $k < 0$, and consequently $a_k = 0$ for $k < 0$, so that $f$ has a removable singularity at $z=P$ as well.