The issue with the underlying monoid seems to complicate everything, in a similar way to how Postnikov decompositions are complicated by the pi1 issue. In that case, the common technique is to fix G=pi1 and consider the Postnikov tower as operating in the category of spaces over K(G,1) rather than in the ordinary category of spaces.
So I don't see a lot of hope immediately for getting the pi0 problem out of the way at the same time; it seems like it colors the whole problem.
Once you've decided on a lifting pi0Btopi0X, though, you can fix the underlying monoid because then you're reduced to studying lifts Btimespi0Ypi0XtoX over pi0X.
If you then fix M=pi0X then there's certainly some kind of obstruction theory, but the problem is identifying the obstruction classes as coming from something cohomological that you can actually calculate. It seems to me that one should study the symmetric monoidal category of "spaces over M", with product having fibers
(XstarY)m=coprodm′m″=m(Xm′timesYm″)
(which is some kind of left Kan extension), and try to get some handle on it.
Even when M=mathbbN the bookkeeping gets complicated. Then you're studying "graded Einfty spaces" and your obstruction theory will land in something like cohomology with coefficients in the relative homotopy groups of Y over B, but you're taking cohomology of the "derived indecomposables" in your Einfty space. The zero'th space of derived indecomposables of an Einfty space B over mathbbN is the topological Andr'e-Quillen homology object of B0. Even if B0 is trivial, then the zero'th derived indecomposable space is trivial, the first is B1, and the next is the homotopy cofiber of the squaring map (B1timesB1)hmathbbZ/2toB2.
Based on this kind of futzing around I am led to believe that your obstructions may possibly occur in the relative topological Andre-Quillen cohomology of Sigmainfty+B over Sigmainfty+M with coefficients in the relative homotopy of X over Y. But the problem seems very difficult for a general monoid M.
(Especially evidenced by the fact that Charles Rezk hasn't popped in here with an answer yet.)
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