It would be much simpler to prove the result considering the family obtained by composing with the Möbius transformation $$z \mapsto \frac{z-1}{z+1}$$ that maps the right half plane biholomorphically to the unit disk.
But well, let's look at what we got from considering $e^{-f}$. Without loss of generality, we can assume that the entire sequence $e^{-f_n}$ converges compactly to a nonzero function $g$.
As you observed, that does not yet guarantee that the sequence $f_n$ itself converges compactly to a holomorphic function. So let's fix some $z_0 \in \Omega$ and consider the sequence $f_n(z_0)$. Either the sequence converges to $\infty$, or we can extract a subsequence converging to a complex number.
Consider first the case where we can extract a subsequence converging to a complex number. Without loss of generality, assume the entire sequence converges to $w_0 \in \mathbb{C}$. In a neighbourhood of $e^{-w_0}$, there is a branch of the logarithm with $\log e^{-w_0} = -w_0$ defined.
Then $f_n$ converges uniformly to $\log g$ in a neighbourhood of $z_0$.
If $f_n(z_0) \to \infty$, then, taking a branch of the logarithm in a neighbourhood of $g(z_0)$, we obtain a sequence $k_n$ of integers with $\lvert k_n\rvert \to \infty$ and $f_n(z_0) - 2\pi i k_n \to \log g(z_0)$. Thus the sequence $f_n - 2\pi i k_n$ converges uniformly to a holomorphic function in a neighbourhood of $z_0$, and since $\lvert k_n\rvert \to \infty$, the sequence $f_n$ itself converges uniformly to $\infty$ in a neighbourhood of $z_0$.
It remains to see that the uniform convergence to either a holomorphic function or $\infty$ extends (as locally uniform convergence) to all of $\Omega$.
Let $A = \{z \in \Omega : f_n(z) \to \infty\}$ and $B = \{z \in \Omega : f_n(z) \text{ is bounded}\}$ and $C = \Omega \setminus (A\cup B)$.
The argument above shows that all, $A$, $B$ and $C$ are open, and they are disjoint. Since $\Omega$ is connected, we have $\Omega = A$, $\Omega = B$, or $\Omega = C$. By having extracted the subsequence converging (to $\infty$ or $w_0$) at $z_0$, we have arranged that $z_0 \notin C$, hence $C = \varnothing$, so $\Omega = A$ if $f_n(z_0) \to \infty$, and $\Omega = B$ if $f_n(z_0) \to w_0$.
Best Answer
$f_n(z)=\frac{z+a_n}{1+\bar a_n z}$ for any $|a_n|<1, a_n \to 1$ will do; take for example $a_n=1-\frac{1}{n}$ ($f_n(0)=a_n \to 1$ so any normal covergent subsequence of $f_n$ must converge to $1$ by maximum modulus, but the family is uniformly bounded so it must converge normally to $1$ as any subsequence has a convergent sub-subsequence)