using $exp(\xi \cdot x)$
$g_{ab} = e_{ab} exp(\xi_u x^u)$
For $e_{ab}$ symmetric and invertible.
$g^{ab} = (e^{-1})^{ab} exp(-\xi_u x^u)$
$g_{ab,c} = e_{ab} \xi_c exp(\xi_u x^u) = g_{ab} \xi_c$
$\Gamma_{abc} = \frac{1}{2} (g_{ab} \xi_c + g_{ac} \xi_b - g_{bc} \xi_a)$
${\Gamma^a}_{bc} = \frac{1}{2}(\delta^a_b \xi_c + \delta^a_c \xi_b - g_{bc} \xi^a)$
${\Gamma^a}_{bc,d} = -\frac{1}{2}( g_{bc,d} \xi^a + g_{bc} {g^{ae}}_{,d} \xi_e)$
$= -\frac{1}{2}( \xi^a g_{bc} \xi_d - g_{bc} g^{af} g_{fg,d} g^{ge} \xi_e)$
$= -\frac{1}{2}( \xi^a g_{bc} \xi_d - g_{bc} g^{af} g_{fg} \xi_d g^{ge} \xi_e)$
$= -\frac{1}{2}( \xi^a g_{bc} \xi_d - g_{bc} \delta^a_g \xi_d g^{ge} \xi_e)$
$= -\frac{1}{2}( \xi^a g_{bc} \xi_d - \xi^a g_{bc} \xi_d)$
$= 0$
${\Gamma^a}_{uc} {\Gamma^u}_{bd} = \frac{1}{4}(\delta^a_u \xi_c + \delta^a_c \xi_u - g_{uc} \xi^a)(\delta^u_b \xi_d + \delta^u_d \xi_b - g_{bd} \xi^u)$
$= \frac{1}{4}(
\delta^a_u \xi_c \delta^u_b \xi_d
+ \delta^a_u \xi_c \delta^u_d \xi_b
- \delta^a_u \xi_c g_{bd} \xi^u
+ \delta^a_c \xi_u \delta^u_b \xi_d
+ \delta^a_c \xi_u \delta^u_d \xi_b
- \delta^a_c \xi_u g_{bd} \xi^u
- g_{uc} \xi^a \delta^u_b \xi_d
- g_{uc} \xi^a \delta^u_d \xi_b
+ g_{uc} \xi^a g_{bd} \xi^u
)$
$= \frac{1}{4}(
\delta^a_b \xi_c \xi_d
+ \delta^a_d \xi_b \xi_c
- \xi^a \xi_c g_{bd}
+ \delta^a_c \xi_b \xi_d
+ \delta^a_c \xi_d \xi_b
- \delta^a_c g_{bd} \xi^u \xi_u
- g_{bc} \xi^a \xi_d
- g_{dc} \xi^a \xi_b
+ \xi^a g_{bd} \xi_c
)$
$= \frac{1}{4}(
+ \delta^a_b \xi_c \xi_d
+ \delta^a_d \xi_b \xi_c
+ 2 \delta^a_c \xi_b \xi_d
- \delta^a_c g_{bd} \xi^u \xi_u
- g_{bc} \xi^a \xi_d
- g_{cd} \xi^a \xi_b
)$
${R^a}_{bcd} = \frac{1}{4}(
+ \delta^a_b \xi_c \xi_d
+ \delta^a_d \xi_b \xi_c
+ 2 \delta^a_c \xi_b \xi_d
- \delta^a_c g_{bd} \xi^u \xi_u
- g_{bc} \xi^a \xi_d
- g_{cd} \xi^a \xi_b
- \delta^a_b \xi_d \xi_c
- \delta^a_c \xi_b \xi_d
- 2 \delta^a_d \xi_b \xi_c
+ \delta^a_d g_{bc} \xi^u \xi_u
+ g_{bd} \xi^a \xi_c
+ g_{cd} \xi^a \xi_b
)$
$= \frac{1}{4}(
- \delta^a_d \xi_b \xi_c
+ \delta^a_c \xi_b \xi_d
- g_{bc} \xi^a \xi_d
+ g_{bd} \xi^a \xi_c
- \delta^a_c g_{bd} \xi^u \xi_u
+ \delta^a_d g_{bc} \xi^u \xi_u
)$
$R_{ab} = \frac{1}{4}(
- \delta^c_b \xi_a \xi_c
+ \delta^c_c \xi_a \xi_b
- g_{ac} \xi^c \xi_b
+ g_{ab} \xi^c \xi_c
- \delta^c_c g_{ab} \xi^u \xi_u
+ \delta^c_b g_{ac} \xi^u \xi_u
)$
$= \frac{1}{4}( 2 \xi_a \xi_b - 2 g_{ab} \xi^u \xi_u)$
$= \frac{1}{2}(g_{ac} g_{bd} - g_{ab} g_{cd}) \xi^c \xi^d)$
$= -g_{a[b} g_{c]d} \xi^c \xi^d)$
$R = \frac{1}{2}(g^{ab} g_{ac} g_{bd} - g^{ab} g_{ab} g_{cd}) \xi^c \xi^d)$
$R = -\frac{3}{2} \xi^u \xi_u$
$R = -\frac{3}{2} \xi^u \xi_u$
$\frac{-2 R}{3} = \xi^u \xi_u$
Therefore if $R = 0$
then $\xi^u \xi_u = 0$
then $\xi_a \xi_b (e^{-1})^{ab} = 0$
then $\xi$ is a null vector with inner product weighted by $(e^{-1})$
Now let's look at the components of the Ricci:
$R_{ab} = \frac{1}{4}( 2 \xi_a \xi_b - 2 g_{ab} \xi^u \xi_u)$
$R_{ab} = \frac{1}{2} \xi_a \xi_b + \frac{1}{3} g_{ab} R$
For $R = 0$ this gives
$2 R_{ab} = \xi_a \xi_b$
So if $R = 0$ then $R_{ab}$ must be the outer product of $\xi$
$G_{ab} = R_{ab} - \frac{1}{2} g_{ab} R$
$= \frac{1}{2} \xi_a \xi_b - \frac{1}{2} g_{ab} \xi^u \xi_u - \frac{1}{2} g_{ab} (-\frac{3}{2} \xi^u \xi_u)$
$= \frac{1}{2} \xi_a \xi_b + \frac{1}{4} g_{ab} \xi^u \xi_u$
EFE vacuum solution for $G_{ab} = 0$
$0 = \frac{1}{2} \xi_a \xi_b + \frac{1}{4} g_{ab} \xi^u \xi_u$
$g_{ab} = -2 \xi_a \xi_b / (\xi^u \xi_u)$
$= -2 \xi_a \xi_b / (\xi_u \xi_v g^{uv})$
Let $g_{ab} = \phi \xi_a \xi_b$
Then ... $g_{ab}$ isn't invertible ... and we can't solve this.
Let's try again from the start.
Let $\phi = exp(\xi_u x^u)$
$g_{ab} = \eta_{ab} + e_a e_b exp(\xi_u x^u) = \eta_{ab} + e_a e_b \phi$
$g = det([g_{ab}]) = -1 - \phi \eta^{ab} e_a e_b$
$g^{ab} = (\phi \eta^{au} e_u \eta^{bv} e_v - \eta^{ab} (1 + \phi e_u \eta^{uv} e_v) ) / g$
...where $\eta_{ab} = \eta^{ab} = diag(-1, 1, 1, 1)$ and all else is raised/lowered by $g_{ab}$ and $g^{ab}$
All the $\Gamma_{abc}$, ${\Gamma^a}_{bc}$, ${R^a}_{bcd}$, $R_{ab}$, $R$, $G_{ab}$ are all the same.
Now the EFE $G_{ab} = 0$ gives us:
$g_{ab} = -2 \xi_a \xi_b / (\xi_u \xi_v g^{uv})$
$= -2 \xi_a \xi_b / (\xi_u \xi_v
(\phi \eta^{uc} e_c \eta^{vd} e_d - \eta^{uv} (1 + \phi e_c \eta^{cd} e_d) ) / g
)$
$= -2 g \xi_a \xi_b / (
\phi \cdot \xi_u \eta^{uc} e_c \cdot \xi_v \eta^{vd} e_d - \xi_u \xi_v \eta^{uv} - \phi \cdot \xi_u \eta^{uv} \xi_v \cdot e_c \eta^{cd} e_d
)$
$= 2 (1 + \phi e_e \eta^{ef} e_f) \xi_a \xi_b / (
\phi \cdot \xi_u \eta^{uc} e_c \cdot \xi_v \eta^{vd} e_d - \xi_u \xi_v \eta^{uv} - \phi \cdot \xi_u \eta^{uv} \xi_v \cdot e_c \eta^{cd} e_d
)$
Let $e_a = \theta \xi_a$
$= 2 (1 + \phi \theta^2 \xi_e \eta^{ef} \xi_f) \xi_a \xi_b / (
\phi \theta^2 \cdot \xi_u \eta^{uc} \xi_c \cdot \xi_v \eta^{vd} \xi_d - \xi_u \xi_v \eta^{uv} - \phi \theta^2 \cdot \xi_u \eta^{uv} \xi_v \cdot \xi_c \eta^{cd} \xi_d
)$
$= -2 (1 / (\xi_u \xi_v \eta^{uv}) + \phi \theta^2) \xi_a \xi_b$
$= -2 / (\xi_u \xi_v \eta^{uv}) - 2 \phi e_a e_b$
...and now we have the problem that the only $\phi e_a e_b$ term is scaled by $-2$...