I think that the assumption that the quantum Lorentz group is not connected is wrong, and this is due to some confusion about "duality", a word used, in this context, with different meanings. This is not special of $SL_2$.
I partly share this confusion and I hope that my message will be corrected by someone with clearer ideas than I have.
The compact group $K$ has a Pontryagin dual $\widehat{K}$. Having fixed on $K$ the standard Poisson-Lie structure, it also has a Poisson dual $K^*$ which is $AN$. the two things are patently not the same.
Now the point is that on one hand the universal enveloping Lie algebra $u(\mathfrak{k})$ may be identified with a quantization of the group $AN$ (this is called quantum duality principle) and on the other hand is related (sorry, here I'm really "handwaving") to $\widehat{K}$ (guess to the convolution algebra).
The keypoint is exactly this last relation, which is the right classical analogue of the quantum case.
The relation between $AN$ and $\widehat{K}$ can be considered here a semiclassical effect.
EDIT:
Say $K$ is a compact Lie group and $R(K)$ its Lie algebra of representative functions. Then $R(K)$ is isomorphic to the group algebra of its Pontryagin dual $\mathbb C[\widehat{K}]$ (this can be seen as a version of Fourier transform).
You may then say that the universal enveloping algebra $U(\mathfrak{k})$ is dual to $\mathbb C[\widehat{K}]$.
On the other hand $U(\mathfrak k)$ is a quantization of the algebra of functions on the dual of the Lie algebra, $\mathfrak k^\ast$, with respect to the linear Lie-Poisson bracket.
So the same object is related on one side to the discrete group $\widehat{K}$ and on the other side to the Poisson manifold $\mathfrak k^\ast$. Here there is no quantum group appearing, since we are assuming the trivial Poisson-Lie bracket on $K$.
In the paper cited above we are assuming the standard Poisson-Lie structure on $K=SU(2)$. This implies that the Poisson manifold $\mathfrak{su}(2)^*$ has to be replaced by $AN$ with the dual Poisson-Lie group structure and that $U(\mathfrak k)$ gets replaced by $U_q(\mathfrak k)$.