The problem in it's entirety is this:
Given some simply connected domain $U$ then show that there is a function $g\in\mathcal{O}(\mathbb{C})$ such that for any given $f$ there exists a sequence $\zeta_{n} \in \mathbb{C}$ so that $g(z-\zeta_{n})$ converges compactly to $f(z)$ in $U$.
I am fairly certain that the problem should be solved in this manner.
There are countably many polynomials with coefficients in $\mathbb{Q}(i)$ therefore we can have a sequence $q_{n}$ where every such polynomial appears infinitely many times. I can show this.
If we have some sequence of non-intersecting balls $V_{n}$ (with centers $\zeta_{n}$) of increasing radius, tending to infinity, then there exists some entire function $g$ so that $\lim_{n\rightarrow \infty}||(g-q_{n})|_{Vn}||=0$. I can not show this.
There exists a sequence of polynomials $p_n$ such that $p_n(z)$ converges compactly to $f(z)$. I can show this.
There exists some subsequence of $q_n$, which we will call $m_{n}$ such that $\lim_{n\rightarrow \infty}\|(p_n-m_n)|_{Vn}\|=0$. I can show this.
Therefore, within any compact, and therefore bounded set we have $|f(z)-g(z-\zeta_{mn})| \leq |f(z)-p_n(z)| + |p_n(z)-m_n(z)| + |m_n(z)-g(z-\zeta_{mn})|$, all of which can be made arbitrarily small with a large enough $n$. I can show this.
I know there are some presentation details I need to fix, however the idea should be clear for now. The main problem lies in being unable to show step #2. If there are any ideas on how to do this please let me know.
The solution to the problem is so tantalizingly close that I feel that this must be the correct approach. Maybe the idea is to choose some clever $V_n$ or use some powerful approximation result that I don't know about.
Perhaps this is somehow related to Montel's theorem?