= Solution
Extract the <Riemann curvature tensor> from $\Theta^a{}_b=\tfrac12R^a{}_{bcd}e^c\wedge e^d$ using the <curvature 2-forms> just computed. Contraction $R_{bd}=R^a{}_{bad}$ gives the diagonal <Ricci tensor> in the <orthonormal coframe>:
$$
R_{00}=q+2a^2,\qquad R_{11}=R_{22}=-(q+2a^2),\qquad R_{33}=-3q,\qquad R=-6q-6a^2.
$$
For example, the $00$ component receives $a^2$ from each of the $1$ and $2$ directions and $q$ from direction $3$. The signs in the spatial components reflect the same Lorentzian index lowering used for the <connection 1 forms>.
For a <null vector> $k$ in this frame, $(k^0)^2=(k^1)^2+(k^2)^2+(k^3)^2$. Therefore
$$
R_{ab}k^ak^b=(q+2a^2)\bigl((k^0)^2-(k^1)^2-(k^2)^2\bigr)-3q(k^3)^2=2(a^2-q)(k^3)^2=-2a'(k^3)^2.
$$
The <Einstein field equations> imply $8\pi T_{ab}k^ak^b=R_{ab}k^ak^b$, since the metric term vanishes for a <null vector>; an included cosmological-constant term would also vanish. Choose $k^a=(1,0,0,1)$ and apply the <null energy condition>. It follows that
$$
\boxed{\frac{d^2}{dz^2}\log A=a'=\frac{A''}A-\frac{(A')^2}{A^2}\le0.}
$$
Conversely, this inequality makes the same contraction nonnegative for every <null vector>, so it is exactly the <null energy condition for a planar warped spacetime> when its matter <stress-energy tensor> is defined by the <Einstein field equations>. Null vectors tangent to the planar slices saturate the condition. As checks, constant $A$ gives flat spacetime and $A=e^{bz}$ gives constant negative sectional curvature with $a'=0$, also saturating it.
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