Does every elliptic cohomology theory represent a complex-orientable $E_\infty$-ring spectra and vice-versa? 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 $K(ku)$?

Neither $K(ku)$ nor tmf are complex orientable, so neither $K(ku)$ nor tmf are elliptic cohomology theories in the strict sense. $K(ku)$ "is" an elliptic cohomology theory in the looser sense that it "detects $v_2$-periodic phenomena" (although I can't elaborate too much on what this means), and tmf "is" an elliptic cohomology theory in the looser sense that it is built out of all elliptic cohomology theories somehow. 
