Is $\mathsf{P} = \mathsf{NP}$ relative to a universal predictor?
I am not familiar with the notion of universal predictor, and I did not follow everything you wrote; in particular, I did not follow your sketch of the proof of existence of a universal predictor in E.  But assuming that there exists a universal open predictor that belongs to E, the answer to your question is positive.  And I am afraid that you will probably be disappointed by the reason.

Edit in revision 2: I changed the construction in response to the added restriction in revision 2 of the question, but the general idea is the same: mix a hard language into a universal open predictor in such a way that the definition of universal open predictor does not notice the difference.

Let T = {0|x|11x: x∈{0,1}*}.  Note that every word in T has an even length.  An important property of T is that no word in T is a proper prefix of another word in T.  This implies that if the symmetric difference between two languages V and W is contained in T, then for every infinite sequence s∈{0,1}ω, there is at most one n such that χV(s&lt;2n)≠χW(s&lt;2n), and in particular V is a universal open predictor if and only if W is a universal open predictor.

It is easy to see that there exists an EXPSPACE-complete language which is a subset of T.  Let L be such a language.  Let V be a universal open predictor that belongs to E, and hence also to EXPSPACE.  Define a language W = L ∪ (V∖T).  Because V is a universal open predictor and the symmetric difference between V and W is contained in T, W is also a universal open predictor.  It is easy to see that W is EXPSPACE-complete, and therefore PW = NPW = EXPSPACE.  This concludes that W satisfies the desired property.