This question has an answer here. To this I would like to add does the function $$f(z)=e^{-z}$$ satisfies the requirement? This is an entire function and $0$ is removable singularity of $f(1/z)$ as $$\lim_{z\to0}e^{-1/z}=0$$ Also extending this, consider non-constant polynomial $g(z)$ the $e^{-g(z)}$ is entire function. $z=0$ is the pole of $g(1/z)$ so $$\lim_{z\to0}e^{g(-1/z)}=0$$
So all functions of the form $e^{-g(z)}$ satisfies these conditions.
Is this correct? Are there any other functions?