The existence of the decomposition does not depend on the assumption on the real part of $f$. The shortest way to get it is to expand $f$ in Laurent series in some neighbourhood of infinity $W=\{ \vert z\vert >a\}$ :
$$f(z)=\sum_{n=-\infty}^{-1} c_n z^n + \sum_{n=0}^\infty c_n z^n\, .$$
The first sum is $f_1(z)$. The second series is a power series converging in $W$, hence in fact in the whole complex plane, so it defines an entire function $f_2(z)$.
If you want to follow the hint to get the decomposition, you can proceed as follows. Fix $R_0$ such that $f$ is holomorphic in an open set containing $\{\vert z\vert\geq R_0\}$. For each $R\geq R_0$, denote by $\Gamma_R$ the circle $\{\vert z\vert=R\}$ and define
$$f_R(z)=\frac{1}{2i\pi}\int_{\Gamma_R} \frac{f(\xi)}{\xi -z}d\xi\, ,$$
which is holomorphic in $\mathbb C\setminus\Gamma_R$. By Cauchy's formula, we have
$$f(z)=f_R(z)-f_{R_0}(z)$$
whenever $R_0<\vert z\vert <R$. This shows that if $R_0<R< R'$, then the functions $f_R$ and $f_{R'}$ agree in $\{ R_0<\vert z\vert<R\}$, and hence on the disk $D(0,R)$ by analytic continuation. In other words (letting $R\to\infty$), we get a single entire function $f_2$ equal to $F_R$ on $D(0,R)$ for any $R>R_0$. Finally, put $f_1=-f_{R_0}$.
Having done this, observe that the function $f_2$ satisfies the same assumption as $f$, since $f$ and obviously $f_1$ do. Moreover, $f_1$ has a limit at $\infty$ (!) Thus, to prove that $f$ has a limit at $\infty$, it is in fact enough to assume that $f$ is an $entire$ function.
At this point, I don't see how to conclude without using the $Borel$-$Caratheodory\;inequality$, which allows to control $\vert f\vert$ by ${\rm Re} f$. A special case of this inequality reads as follows : for any $r>0$
$$\sup_{\vert z\vert=r} \vert f(z)\vert\leq 2\sup_{\vert z\vert=2r} {\rm Re}f(z)+3\vert f(0)\vert\, .$$
From this, it follows that $z^{-1} f(z)\to 0$ as $z\to\infty$. Then the usual proof of Liouville's theorem shows that $f$ is constant (recall that we are assuming that $f$ is entire).
First question: A function is (complex) analytic if and only if it is holomorphic. It is an easy exercise to see that if $f$ is complex differentiable in $w$ and $f(w) \neq 0$, then $1/f$ is also complex differentiable in $w$ and $(1/f)'(w) = -f'(w)/f(w)^2$. So we know that $1/f$ is holomorphic in $B(z_0,\delta)\setminus \{z_0\}$, hence analytic.
Second question: It is not correct that $z_0$ is an isolated singularity of $1/f$ since $\lim\limits_{z\to z_0} \frac{1}{f(z)} = 0$. It is an isolated singularity since $1/f$ is holomorphic in $B(z_0,\delta)\setminus\{z_0\}$. The limit says that $z_0$ is a removable singularity, and the value which removes the singularity is $0$, so
$$g(z) = \begin{cases}\frac{1}{f(z)} &, z \neq z_0\\ 0 &, z = z_0 \end{cases}$$
is holomorphic on $B(z_0,\delta)$.
Third question: Since $g(z_0) = 0$, the power series representation of $g$ about $z_0$ has a constant term zero, hence
$$g(z) = \sum_{k=1}^\infty a_k (z-z_0)^k.$$
since $g$ does not vanish identically, not all coefficients are $0$, and if we let $m = \min \{ k \in\mathbb{N} : a_k \neq 0\}$, we have
$$g(z) = \sum_{k=m}^\infty a_k (z-z_0)^k$$
with $a_m \neq 0$. Then
$$h(z) = \sum_{k=m}^\infty a_k (z-z_0)^{k-m} = \sum_{r=0}^\infty a_{r+m}(z-z_0)^r$$
is holomorphic in $B(z_0,\delta)$ with $h(z_0) = a_m \neq 0$, and evidently $g(z) = (z-z_0)^m\cdot h(z)$. Since $g$ has no zero in $B(z_0,\delta)\setminus\{z_0\}$, $h$ does not vanish anywhere in $B(z_0,\delta)$, so $\tilde{h}(z) = \frac{1}{h(z)}$ is holomorphic on $B(z_0,\delta)$, and we have
$$f(z) = (z-z_0)^{-m}\cdot \tilde{h}(z)$$
on $B(z_0,\delta)\setminus\{z_0\}$.
Best Answer
Update:
Assume that ${\rm Im}\bigl(f(z)\bigr)<M$ for some $M>0$ and all $z\in\dot\Omega:=\Omega\setminus\{a\}$. The Moebius transform $$w\mapsto \zeta:={iMw\over 2iM-w}$$ maps the half plane ${\rm Im}(w)<M$ onto the interior of the disc $|\zeta|<M$. It follows that the function $$g(z):={iM f(z)\over 2iM-f(z)}$$ satisfies $$|g(z)|<M\qquad\forall z\in\dot\Omega\ .$$ Therefore $g$ has a removable singularity at $a$ and is in fact analytic in all of $\Omega$. By the maximum principle it follows that $|g(a)|<M$ as well, whence $$f(z)={2iM g(z)\over iM+g(z)}$$ is analytic in all of $\Omega$.