The "entire finite complex plane" is just the complex plane. This phrase isn't used very often, but can be used if you want to differentiate between the complex plane $\mathbb{C}$ and the Riemann sphere $\mathbb{C}\cup \{\infty\}$.
The easiest way to answer this is to use Little Picard, but there are certainly easier ways using less powerful machinery. For example, you could consider $\exp(if(z))$ and use Liouville.
Edit: Since you've added the tag reference-request, I will mention that the term "finite complex plane" is used in Silverman's translation of Markushevich's monumental "Theory of Functions of a Complex Variable" which is one of the standard references in complex analysis.
I am writing, enlarging and enhancing (hopefully...) Mex's answer to his own question (kudos!) and I'll be happy to erase this answer of mine if he decides to write down his.
We have that $$f(z)=\sum_{n=0}^\infty a_nz^n$$ because $\,f\,$ is entire, and by the given conditions we have$$(1)\,\,f(r)=\sum_{n=0}^\infty a_nr^n\in\mathbb{R}\,,\,\,\,r\in\mathbb{R}$$$$(2)\,\,f(ir)=\sum_{n=0}^\infty a_n(ir)^n\in i\mathbb{R}\,\,,\,\,r\in\mathbb{R}$$but we have that $$\sum_{n=0}^\infty a_n(ir)^n=\sum_{n=0}^\infty i^n (a_nr^n)=\sum_{n=0}^\infty (-1)^na_{2n}r^{2n}+i\sum_{n=0}^\infty (-1)^na_{2n+1}r^{2n+1} $$and as the above is purely imaginary we get that $\,a_{2n}=0\,,\,\forall n\in\mathbb{N}\,$ , so the power series of the function has zero coefficients for the even powers of $\,z\,$ and is thus a sum of odd powers and trivially then an odd function.
Best Answer
Let $f_1$ be the restriction of $f$ to $\{\Im(z)\ge 0\}$. By Schwarz's reflection, this admits an analytic extension to $\mathbb{C}$ defined in $\Im(z)\le 0$ by $g(z)=\overline{f_1(\bar z)}$. By the identity principle, $g=f$. Since $f$ is purely imaginary on $i\mathbb{R}$, this equation tells us that $f(z)+f(-z)$ is zero on $i\mathbb{R}$. By the identity principle again, $f(z)+f(-z)\equiv 0$, which implies that $f$ is odd.