Your calculation is correct. In Hatcher's description of integral cohomology mod torsion the last generator has to be interpreted as an extra generator in addition to the previous ones.

The integral cohomology ring of the limit is in general not isomorphic to the E_{\infty} ring, but the E_{\infty} ring is the associated graded to a filtration on the limit.

You might want to look at the Leray-Hirsch theorem which tells you about part of the cup product structure whenever the fundamental group of the base acts trivially and there are no differentials. (And you can use that the induced maps H^(B)\to H^(E)\to H^*(F) in cohomology are ring isomorphisms.)

Anyway, with all of these theorems I am still missing one cup product H^2 \times H^3 \to H^5.

Maybe one has to use what is described in Hatcher's note?

Your calculation is correct. In Hatcher's description of integral cohomology mod torsion the last generator has to be interpreted as an extra generator in addition to the previous ones.

The integral cohomology ring of the limit is in general not isomorphic to the E_{\infty} ring, but the E_{\infty} ring is the associated graded to a filtration on the limit.

You might want to look at the Leray-Hirsch theorem which tells you about part of the cup product structure whenever the fundamental group of the base acts trivially and there are no differentials. (And you can use that the induced maps H^(B)\to H^(E)\to H^*(F) in cohomology are ring isomorphisms.)

In your particular case it turns out that there is only one possibility: the E_{\infty} and the limit are isomorphic as rings.