The vielbein can be recovered from the curvature (see this post for example).
Here I want to show an even simpler way of doing this and to show, further, that the space of vielbeins satisfying some conditions is bijective with curvature 2-forms themselves satisfying some conditions. The motivation for this is that I wanted to generalize a result known in gauge theories -- that there is a bijection between connections and curvatures (see https://journals.aps.org/prd/abstract/10.1103/PhysRevD.19.517 for example) -- to the case of metrics and curvatures.
Question: Are the following claims and proofs all correct?
1) Basics
We will use Greek indices in the coordinate basis and Latin indices in the vielbein basis. $e^a$ denotes the vielbein $1$-forms and $e_a$ their vector duals. The spin connection and curvature $2$-form are defined as \begin{align} \omega^a_b &= \frac{1}{2}(-e^c\wedge i_bi_ade^c+i_bde^a-i_ade^b) \tag{1} \\ R^a_b &= d\omega^a_b+\omega^a_c\wedge \omega^c_b \tag{2} \end{align} where $i_a$ is the interior product w.r.t. $e_a$. Using the fact that $e^b\wedge i_b(\eta)=p\eta$ for a $p$-form $\eta$, we can show (1) is equivalent to \begin{equation} de^a+\omega^a_b\wedge e^b=0 \tag{3} \end{equation} Let $r\equiv x^\mu\partial_\mu$. Since $[r,\partial_\mu]=-\partial_\mu$ we have $$ (\mathcal L_r \eta)_{\mu_1...\mu_p}=(r+p)\eta_{\mu_1...\mu_p} \tag{4} $$ where $\eta$ is any $p$-form.
2) Going between the curvature and connection
Assume the gauge condition $i_r\omega^a_{b}=0$ (which fixes the $SO(d)$ gauge freedom completely). Since $\mathcal L_r =\{d,i_r\}$ we have $$ i_r R^a_b= \mathcal L_r\omega^a_b \tag{5} $$ Therefore \begin{equation} x^\mu R^a_{b\mu\nu}= (r+1)\omega^a_{vb} \tag{6} \end{equation} which is easily integrated to give \begin{equation} \omega^a_{\nu b}(x)=\int_0^1 tx^\mu R^a_{b\mu\nu}(tx) dt \tag{7} \end{equation} Notice that $i_r\omega^a_b=0$ automatically from this.
3) Claim: curvatures satisfying the 2nd Bianchi identity are bijective with connections satisfying the gauge condition
Consider the space of curvatures satisfying the 2nd Bianchi identity $dR^a_b+\omega^a_c\wedge R^c_b-R^a_c\wedge \omega^c_b=0$ where $\omega$ in this expression is given by (7). Taking the interior product and using (5) gives $$ i_r(dR^a_b+\omega^a_c\wedge R^c_b-R^a_c\wedge \omega^c_b)=\mathcal L_r(R^a_b-d\omega^a_b-\omega^a_c\wedge \omega^c_b )=0 \tag{8} $$ which implies $R^a_b=d\omega^a_b+\omega^a_c\wedge \omega^c_b$, with $\omega$ in this expression given by (7). This proves the bijection.
4) Going between the vielbein and connection
Taking the interior product of (3) gives $$ -di_re^a+(r+1) e^a-\omega^a_b \wedge i_r e^b=0 \tag{10} $$ Assume we are working in normal coordinates such that $x^\mu g_{\mu\nu}=x^\nu$ -- which implies $x^\mu e^a_\mu=x^a$ and $x^a e^a_\mu=x^\mu$. Therefore (10) becomes $$ (r+1)e^a_\nu=\delta^a_\nu+\omega^a_{\nu b}x^b \tag{11} $$ which is easily integrated to give $$ e^a_\nu=\delta^a_\nu+\int_0^1 \omega^a_{\nu b}(tx) tx^b dt \tag{12} $$ Using (7) gives $$ e^a_\nu=\delta^a_\nu+\int_0^1\int_0^1 t_1t_2^2x^b x^\mu R^a_{b\mu\nu}(t_1t_2x) dt_1 dt_2 \tag{13} $$ Fixing $t_1t_2$ and doing the $t_2$ integral first shows that $$ e^a_\nu=\delta^a_\nu+\int_0^1 t(1-t)x^b x^\mu R^a_{b\mu\nu}(tx) dt \tag{14} $$ Note that $x^\nu e^a_\nu=x^a$ and $x^a e^a_\nu=x^\nu$ automatically.
5) Claim: curvatures satisfying the 1st and 2nd Bianchi identities are bijective with connections satisfying the gauge and torsionless conditions
We proved in 3) that the space of curvatures satisfying the 2nd Bianchi identity is in bijection with the space of connections satisfying $i_r\omega^a_b=0$. We now place the 1st Bianchi identity $R^a_b\wedge e^b=0$ -- with $e$ given by (14) -- as an additional restriction on the space of curvatures. Via the bijection this descends into an extra restriction on the space of connections: namely $R^a_b\wedge e^b=0$, with $R$ given by (2) and $e$ by (12). Taking the interior product of this gives: $$ i_r(R^a_b\wedge e^b)=\mathcal L_r(de^a+\omega^a_b \wedge e^b)=0. \tag{15} $$ which implies $de^a+\omega^a_b \wedge e^b=0$ with $e$ in this expression given by (12) (we will refer to this as the torsionless condition). Conversely, taking the exterior derivative of the torsionless condition gives the 1st Bianchi identity. Taken together, the bijection is proved.
6) Claim: connections satisfying the gauge and torsionless conditions are bijective with vielbeins satisfying (16)-(18)
Consider the space of vielbeins satisfying \begin{align} x^\mu e^a_\mu&=x^a \tag{16} \\ x^a e^a_\mu&=x^\mu \tag{17} \\ i_a \mathcal L_r e^b&=i_b \mathcal L_r e^a \tag{18} \end{align} Connections arising from such vielbeins satisfy the gauge and torsionless conditions automatically. Conversely, consider the image of connections satisfying the gauge and torsionless conditions under the map (12). It is clear that (16)-(17) hold automatically. On the other hand, as shown in 1), the torsionless condition is equivalent to (1). Applying $i_r$ to (1) and using $x^c \wedge e^c = x^\mu dx^\mu$ gives (18). This proves the bijection.
Combining 5) and 6) proves that the space of curvatures satisfying the 1st and 2nd Bianchi identities is bijective with the space of vielbeins satisfying (16)-(18).
Finally a remark on the Riemann metric and its relation to the vielbeins. The metric is expressed as $g_{ij}=e^a_ie^a_j$. It is clear then how to get the metric from the vielbein. But how does one get the vielbein from the metric? Differentiating the expression for the metric and using (18) (this condition completely fixes the $SO(d)$ gauge freedom the vielbeins have) shows that $$ \frac{d}{dt} e^b_j(tx)=\frac{1}{2}e^i_b(tx) \frac{d}{dt} g_{ij}(tx) \tag{19} $$ Given the metric, this can be integrated uniquely for the vielbeins using the initial condition $e^b_j(0)=\delta^b_j$.