I want to prove the following result: Let $\Omega$ be the domain $\mathbb{C}\backslash \{a_1,a_2,a_3,a_4\}$. Its universal covering is the unit disk, and the standard Poincaré metric is pulled back to a metric $ds=\rho(z)|dz| $ on $\Omega$. For $\rho(z)$ we have the asymptotic expansions $$\rho(z)\sim \frac{C_j}{|z-a_j|\log|z-a_j|} \text{as}\ z\rightarrow a_j$$ and $$\rho(z)\sim \frac{C_0}{|z|\log|z|} \text{as}\ z\rightarrow \infty$$ where $C_j$ are constants different from zero.
Is it possible to give an elementary proof, accessible to someone who has little knowledge on Riemann surfaces?
On the existence of Poincare metric.
There is no really simpler proof for the complement of $5$ points than the general proof that applies to all plane domains. One of the simplest proofs for arbitrary plane domains is here:
MR0962807 Y. Fisher, J. Hubbard, B. Wittner, A proof of the uniformization theorem for arbitrary plane domains. Proc. Amer. Math. Soc. 104 (1988), no. 2, 413–418.
If the $5$ points are on the real line, one can give a simpler proof, which involves a construction of the hyperbolic pentagon with all angles zero and prescribed moduli. That there exists such a pentagon with prescribed moduli is proved by continuity argument. Then you map it on the upper half-plane, so that the vertices go to your $a_j$ and get the universal covering. I doubt that this continuity argument, if done carefully is really simpler than the proof in the general case. (If would be simpler if you had 4 punctures rather than 5).
On the asymptotics.
The standard way to do this is Comparison Principle for the hyperbolic metric. The universal covering of the punctured disk is explicitly known, it is $e^z$ restricted to the left half-plane. So the hyperbolic metric can be explicitly computed for this case. Near a puncture of any region it behaves the same way, by comparison principle. You can also compare with the hyperbolic metric of the sphere minus $3$ points which is also explicitly known.