I think the standard counterexample is this: $\newcommand{\eps}{\varepsilon}$
Let $\Omega_\eps = \{ (z,w) : |z| < 1, |w| < 1, |zw| < \eps \}$ and let $P = (0,1/2)$, $Q = (1/2, 0)$.
Then (using the mapping $f(\zeta) = (0,\zeta)$), it follows that $d_{\Omega_\eps}(P,0) \le \rho(1/2,0)$. Similarly $d_{\Omega_\eps}(0,Q) \le \rho(1/2,0)$.
On the other hand, I claim that $d_{\Omega_\eps}(P,Q) \to \infty$ as $\eps \to 0$. Assume that it does not. Then there is a constant $M$ and a sequence $\eps_j \to 0$ and mappings $f_j = (g_j,h_j) : D \to \Omega_j$ with $f_j(0) = (1/2, 0)$ and $f_j(\sigma_j) = (0,1/2)$ such that $\rho(0, \sigma_j) = \tanh^{-1}|\sigma_j| \le M$.
By Montel's theorem, the components $g_j$ and $h_j$ converge locally uniformly (after passing to a subsequence if necessary) to $g$ and $h$ respectively. By passing to a further subseqence, we may also assume that $\sigma_j \to \sigma$ for some $\sigma$ with $|\sigma| < 1$.
Now, $\| g_j h_j \| < \eps_j$, so in fact $g_j h_j$ converges to $0$ locally uniformly. Hence by the identity principle, either $g$ or $h$ is identically $0$. However, $g_j(0) = 1/2$ for all $j$, so $g$ cannot be identically zero. Also $h_j(\sigma_j) = 1/2$, so $h(\sigma) = 1/2$ and $h$ also cannot be identically $0$.
This contradiction shows that for $\eps$ small enough, $d_{\Omega_\eps}$ does not satisfy the triangle inequality.
