= Solution
For <longitudinal-vector annihilation through a scalar contact interaction>, first use the \b[supplied vertex] $2ig^2\eta_{\mu\nu}$. Let $s=(p_1+p_2)^2$ and work in the <center-of-mass frame>. The amplitude is $\mathcal M=2g^2\epsilon_1\cdot\epsilon_2$, up to its overall phase. For two longitudinal incoming vectors,
$$
\epsilon_L(p_1)\cdot\epsilon_L(p_2)=\frac{p_1\cdot p_2}{m_A^2}=\frac{s-2m_A^2}{2m_A^2},
\qquad
|\mathcal M_{LL}|^2=g^4\frac{(s-2m_A^2)^2}{m_A^4}.
$$
This grows as $s^2/m_A^4$, whereas transverse amplitudes stay of order $g^2$ and mixed longitudinal-transverse amplitudes vanish in this frame. Consequently the leading high-energy contribution is longitudinal.
For an \b[<unpolarized cross section>[unpolarized]] total <cross section>, average over the $3\times3$ incoming <polarizations>. The massive-vector <polarization> sum gives
$$
\sum_{i,j}|\epsilon_i(p_1)\cdot\epsilon_j(p_2)|^2
=2+\frac{(p_1\cdot p_2)^2}{m_A^4}.
$$
The identical final scalars require a factor $1/2!$ in the two-body <phase space>. With $\beta_A=\sqrt{1-4m_A^2/s}$ and $\beta_h=\sqrt{1-4m_h^2/s}$, the angle-independent result is
$$
\boxed{\sigma_{\mathrm{unpol}}(s)=\frac{\beta_h}{\beta_A}\frac{g^4}{288\pi s}
\left[\frac{(s-2m_A^2)^2}{m_A^4}+8\right]
\underset{s\gg m^2}{\sim}\frac{g^4s}{288\pi m_A^4}.}
$$
For a prepared longitudinal pair there is no incoming <polarization> average, giving instead
$$
\boxed{\sigma_{LL}(s)\sim\frac{g^4s}{32\pi m_A^4}=9\sigma_{\mathrm{unpol}}(s).}
$$
There is no extra initial-state factor $1/2$: a <cross section> uses the incoming beam flux, while the final-state factor removes double counting of indistinguishable outgoing states.
Define the <energy> in the stated bound to be the total <center-of-mass energy>, $E=\sqrt s$. Equating the leading expressions to $4\pi/E^2$ gives
$$
\boxed{E_{\mathrm{unpol}}=(1152\pi^2)^{1/4}\frac{m_A}{g}\simeq2.58\ \mathrm{TeV},\qquad
E_{LL}=(128\pi^2)^{1/4}\frac{m_A}{g}\simeq1.49\ \mathrm{TeV},}
$$
for $m_A=100\ \mathrm{GeV}$ and $g=0.4$. The second value shows that a longitudinal channel reaches the supplied perturbative bound earlier. If the source's unspecified $E$ denotes the <energy> of \i[one] incoming particle and the bound is literally $4\pi/E^2$, substitute $s=4E^2$ instead: the corresponding values are $1.83\ \mathrm{TeV}$ <unpolarized cross section>[unpolarized] and $1.05\ \mathrm{TeV}$ longitudinal. Stating the <energy> and <polarization> conventions is necessary for a numerical answer.
\b[Interaction-normalization qualification:] the written $\mathcal L_{\mathrm{int}}=g^2 A_\mu A^\mu h^2$ differentiates to $4ig^2\eta_{\mu\nu}$, because there are $2!$ vector assignments and $2!$ scalar assignments. For example, differentiating $g^2(A^0)^2h^2$ twice in each variable gives $4g^2$. The supplied $2ig^2\eta_{\mu\nu}$ would instead follow from $\mathcal L_{\mathrm{int}}=g^2 A_\mu A^\mu h^2/2$. Thus all <cross sections> above follow the explicitly requested rule. Using the literal written interaction multiplies them by four and divides every quoted threshold <energy> by $\sqrt2$. The <energy> growth itself is the physical issue: a contact interaction of massive longitudinal vectors needs additional dynamics before its perturbative high-energy extrapolation becomes reliable.
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