What does Weinberg–Witten theorem want to express?
The stress tensor for $h_{ab}$ is not Lorentz covariant, despite the fact that it looks like it is. This is because $h_{ab}$ itself is not a Lorentz tensor. Rather under Lorentz transformations $$ h_{ab} \to \Lambda_a{}^c \Lambda_b{}^d h_{cd} + \partial_a \zeta_b + \partial_b \zeta_a ~. $$ The extra term is present to make up for the fact that $h_{ab}$ is not a tensor of the Lorentz group. Plug this into the stress tensor and you will find that the stress tensor also transforms with a inhomogeneous piece thereby making it non-covariant.
The photon is not charged under the $U(1)$ gauge symmetry. Thus, its $U(1)$ current is zero. The current you have defined is not the $U(1)$ current. Rather it is the current corresponding to translations. Weinberg-Witten theorem has nothing to say about this current.
The Weinberg-Witten theorem implies that the graviton is not composite, because quantum fields usually have Lorentz-covariant stress-energy, and composite particles made from such fields will also have Lorentz-covariant stress-energy.
There is an interesting note in the Weinberg-Witten paper that the theorem does not exclude emergent gravity approaches like Sakharov's, because there the emergence is from quantum corrections, and not from composite particles.
Pauli-Fierz theory does not violate the Weinberg-Witten theorem because the stress-energy tensor you constructed is not gauge invariant under the infinitesimal transformation $h_{\mu\nu}\mapsto h_{\mu\nu}+\partial_\mu\chi_\nu + \partial_\nu\chi_\mu$ (this is the Lie derivative $\mathcal{L}_\xi h$ of the metric along the vector field $\chi$, i.e. the infinitesimal change of $h$ under the diffeomorphism generated by $\chi$) for $\chi$ any vector field/1-form. Your $T$ transforms as $$ \delta T_{ab} = 2\partial_a\partial_c\chi_d\partial^c\partial^d\chi_b-\frac{1}{2}\partial_a h \partial^c\partial_c \chi_b - \partial_a\partial_b \chi_c\partial^c h$$ (unless I made a mistake), which is non-zero even after dropping terms second order in $\chi$. Therefore, $T$ is not a gauge invariant stress-energy, and therefore does not violate the Weinberg-Witten theorem.
The $\mathrm{U}(1)$ massless gauge field is not charged under the $\mathrm{U}(1)$ since the adjoint representation of $\mathrm{U}(1)$ is trivial, and hence does not violate the Weinberg-Witten theorem since this theorem explicitly states that the massless field of spin 1 that is forbidden by it is assumed to be charged under the symmetry responsible for the conserved current.