The last paragraph in Two-Vector Bundles and Forms of Elliptic Cohomology remarks that neither the spectrum $K(ku)$ nor tmf is complex orientable. In the case of $K(ku)$: "...the unit map for $K(ku)$ cannot factor through the complex bordism spectrum $MU$, since $\pi_1(MU)=0$."
This confuses me, is it not the case that every elliptic cohomology theory represents a complex-orientable $E_\infty$-ring spectra and vice-versa?
On page 21 of Lurie's Survey, he mentions:
If we view $\mathbb{CP}^\infty$ as the classifying space for complex line bundles, then the group algebra $\Sigma[\mathbb{CP}^\infty]$ can be viewed as a universal cohomology theory in which it is possible to add line bundles. The above result can be viewed as saying that if we take this universal cohomology theory and invert the Bott element $\beta$ then we obtain a theory which classifies vector bundles. A very puzzling feature of the result is the apparent absence of any direct connection of the theory of vector bundles with the problem of orienting the multiplicative group.
Does this mean we have no functorial, multiplicative choice of Thom classes for complex vector bundles in tmf and $K(ku)$?