Suppose $f$ is a function which is holomorphic on $\mathbb{C}\setminus A$ where $A$ is the set of points where $f$ has a singularity. Suppose that all of the points in $A$ are removable singularities of $f$. Here is my question: does this imply that $f$ is itself entire? I understand that by Riemann Extension Theorem, $f$ can be extended to an entire $F$, but my question has to do with whether we can say that $f$ itself is entire. I have seen a few other questions on this site which make such a statement, e.g. Removable singularities and an entire function and I am not sure if they are just slurring notation or if I am missing something.
The context in which this arose: I am trying to show that if two entire functions $f,g$ are such that $|f|\leq |g|$, then one is a multiple of the other. Obviously the strategy is to take quotient, and show that each singularity is removable. I have been able to do this, but then after that I am lost. I know I am supposed to use liouville to show that bounded and entire implies constant, but I am not sure if $|f|/|g|$ is itself entire. Is it not supposed to be some extended function which is supposed to be entire? With such an extended function, indeed we would have bounded and entire, but then I am not sure how to show that $f$ and $g$ are multiples of one another on all of $\mathbb{C}$, since things get strange around the singularities.
I would appreciate anything which clarifies my understanding.