I'm not quite sure what your question is. The U(1) does not rephase SU(5) quintuplets uniformly: it rephases the doublet direct summand differently from the triplet summand, because these have different hypercharges.
Taking generators τ in su(2) to be your 2×2 matrices, and likewise their exponentials, and T in su(3) to be 3×3 matrices, the hypercharge U(1) generator Y is a 5×5 diagonal matrix diag(3,3,-2,-2,-2), suitably normalized. So it commutes with anything in the 2×2 block and anything in the 3×3 block, and is traceless, as a generator of SU(5), but it straddles both blocks, and rephases them differently, albeit in tandem.
That is, Y is not in a separate direct summand, but, as τ, T, and Y all commute amongst themselves, they generate the Cartesian product of the subgroup, i.e., their group parameters are fully independent of each other. Y amounts to two, related, rephasings for the 2 and 3 direct summands of the 5-vectors, respectively.
So, a group element in the unbroken subgroup of SU(5) you are considering is
$$\large
e^{i (\theta \cdot \tau +3\alpha I_2 )~\oplus ~ i(\phi\cdot T -2\alpha I_3)} =e^{i (\theta \cdot \tau +3\alpha I_2) } ~\oplus ~ e^{i(\phi\cdot T -2\alpha I_3)},
$$
where θ, φ, α are the SU(2), SU(3), and hypercharge "angle parameters", respectively (a triplet, octet, and singlet, respectively). Work out the exponential of the block matrices to see how the blocks in the exponent evaluate to a direct sum of blocks as well, perhaps counterintuitively, but pursuant to the direct sum homomorphism, preserving the block split. The first direct summand now acts on the 2×2 block, while the second on the 3×3 block.
It is a neat feature of this unification that the charges of colored particles (the triplet block) are thus pegged to the charges of the leptons ("charge quantization")!