= Solution
The displayed $e^a$ are one-forms, so they form an <orthonormal coframe>, dual to the <orthonormal frame> $E_0=A^{-1}\partial_t$, $E_1=A^{-1}\partial_x$, $E_2=A^{-1}\partial_y$, $E_3=\partial_z$. Use $\eta_{ab}=\operatorname{diag}(-1,1,1,1)$ and abbreviate
$$
a=\frac{A'}A,\qquad q=\frac{A''}A=a'+a^2.
$$
For $i=0,1,2$, $de^i=a e^3\wedge e^i$, while $de^3=0$. The <Levi-Civita connection> is fixed by <Cartan's first structure equation> together with <Lorentzian connection-form antisymmetry>, $\omega_{ab}=-\omega_{ba}$, where $\omega_{ab}=\eta_{ac}\omega^c{}_b$. Its nonzero <connection 1 forms> are
$$
\boxed{\omega^0{}_3=a e^0,\quad\omega^1{}_3=a e^1,\quad\omega^2{}_3=a e^2,\quad\omega^3{}_0=a e^0,\quad\omega^3{}_1=-a e^1,\quad\omega^3{}_2=-a e^2.}
$$
In particular, the time-index pair has the same sign with one index raised; incorrectly imposing $\omega^a{}_b=-\omega^b{}_a$ would lose the Lorentzian sign. All diagonal and all other entries vanish.
Apply <Cartan's second structure equation>. For $i=0,1,2$,
$$
\Theta^i{}_3=d(a e^i)=(a'+a^2)e^3\wedge e^i=-q e^i\wedge e^3.
$$
For distinct $i,j$ in $\{0,1,2\}$, putting $s_0=-1$ and $s_1=s_2=1$ gives $\Theta^i{}_j=\omega^i{}_3\wedge\omega^3{}_j=-s_j a^2e^i\wedge e^j$. Thus a complete independent set of <curvature 2-forms> is
$$
\boxed{\begin{aligned}\Theta^0{}_1&=-a^2e^0\wedge e^1,&\Theta^0{}_2&=-a^2e^0\wedge e^2,&\Theta^1{}_2&=-a^2e^1\wedge e^2,\\\Theta^0{}_3&=-q e^0\wedge e^3,&\Theta^1{}_3&=-q e^1\wedge e^3,&\Theta^2{}_3&=-q e^2\wedge e^3.\end{aligned}}
$$
All diagonal entries vanish. The remaining entries follow from $\Theta_{ab}=-\Theta_{ba}$: $\Theta^j{}_i=-s_is_j\Theta^i{}_j$ for $i,j\le2$, and $\Theta^3{}_i=-s_i\Theta^i{}_3$. These relations and the displayed six entries specify all sixteen <curvature 2-forms>. This computation is <Cartan curvature of a planar warped spacetime>.
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