Q:

Is a tetrad equivalent to a local (position-dependent) Lorentz transformation $e_{\mu}^{a}=\Lambda(x)_{\mu}^{a}$?

The equation in the title won't make since till halfway down.   Let us suppose we are looking at some nonflat spacetime equipped with a metric $g_{\mu\nu}$. Furthermore our manifold admits a covering set of tetrads (vielbeins/framefields whatever you want to call them) such that we can define a local “lab frame”.
$$\eta_{ab}=e_{a}^{\mu}g_{\mu\nu}e_{b}^{\nu}$$
I'm curious here about two different points on a manifold. Since each local frame is Minkowskian , It makes sense to say that any two tetrad frames (say at points $A$ and $B$) can be related by a Lorentz transformation:
$$\eta(B)_{cd}=\Lambda_{c}^{a}e_{a}^{\mu}(A)g_{\mu\nu}(A)e_{b}^{\nu}(A)\Lambda_{d}^{b}$$
The fact that the Minkowski metric is unchanged is hardly suprising; however, our $\Lambda$s though now have to be a function of spacetime positions since different positions will be related to point $A$'s frame by different Lorentz transformations (whereas normally one uses a “global” Lorentz transformation). We could do the same thing with some vector $V^{\mu}$.
$$V(B)^{b}=\left(\Lambda(x^{i})\right)_{a}^{b}e_{\mu}^{a}(A)V^{\mu}$$
$$e_{\mu}^{a}(x_{B})=\Lambda_{b}^{a}(x_{B})e_{\mu}^{b}$$
Suppose there is one finite region of our spacetime that appears Minkowskian, Then the tetrad would simply be the identity matrix within this region. However, we can still represent the tetrad in any other point by the position dependent Lorentz transformation of this frame. Therefore it would appear that:
$$e_{\mu}^{a}=\Lambda(x)_{b}^{a}\mathbf{1}_{\mu}^{b}=\Lambda(x)_{\mu}^{a}$$
$$g_{\mu\nu}=\Lambda(x)_{\mu}^{a}\eta_{ab}\Lambda(x)_{\nu}^{b}$$
Which implies we can represent the tetrad and thus the gravitational field by a position dependent Lorentz transformation! 
It's probably obvious to the reader that there is no way to uniquely define such a transformation between points, which gives the whole setup a large degree of arbitrariness. In particular one could follow two different paths between the points and one should very well expect to obtain two differing $\Lambda$'s. So the whole setup is then defined only up to an overall global Lorentz transformation.
Rather than condemning our construction, this should be of great comfort! It means that we can locally annihilate $\Lambda(x)$ at any point, (and thus the gravitational field) by a choice of global Lorentz transformation. (very much in keeping with the equivalence principle)
Note that since we are using the Lorentz group (a Lie group), we know we can apply Lorentz transformations in the form of rotations (usually denoted $\overrightarrow{J}$) and boosts (denoted $\overrightarrow{K}$). This allows us to write our general transformations as (I will omit constant factors just to unclutter notation) :
$$\Lambda(x)=e^{i\overrightarrow{J}\cdot\overrightarrow{\theta(x)}+i\overrightarrow{K}\cdot\overrightarrow{\phi(x)}}$$
I thought it was pretty interesting. If you wanted to extend this to properly moving fields and whatnot, I imagine you'd want to extend the transformation to the whole of Poincare group.
Note I've swept any talk about conformal changes to the metric under the rug, but I was curious to get another opinion about this setup before I take it to conformal changes. Did I screw this up? 
EDIT: I realize this setup might only apply to a family of geodesic observers (defined everywhere)
ADDENDUM: I think I need to give an example: 
Suppose we push a test person out of the international space station and watch them fall to earth (supposing perfect vacuum all the way down). We could write their local frame at any position as a Lorentz transformation of our own frame on the ISS. mapping them all we could write the general Lorentz transformation as a function of position of the "test person" $\Lambda_{c}^{a}(\overrightarrow{r})$.
$$\Lambda_{c}^{a}(\overrightarrow{r})\left\{ e_{a}^{\mu}\right\} _{ISS}=\left\{ e_{c}^{\mu}\right\} _{TestPerson}$$
Since we could take the ISS to be arbitarily far from the earth, we could let $\left\{ e_{a}^{\mu}\right\} _{ISS}\longrightarrow\mathbf{1}_{a}^{\mu}$ very far away (where $\mathbf{1}_{a}^{\mu}$ is just the identity), which leaves us with simply:
$$\Lambda_{c}^{a}(\overrightarrow{r})\left\{ \mathbf{1}_{a}^{\mu}\right\} _{ISS}=\Lambda_{c}^{\mu}(\overrightarrow{r})=\left\{ e_{c}^{\mu}\right\} _{TestPerson}$$
Written as a Lie group element (as above), the respective rotations and boosts would be a function of position. I think then this contains all of the information of our original tetrad (though it won't in general cases). I think this is just the equivalent of going from a global gauge transformation to a local one, if my minds working right. 
Here we're going from a global lorentz transformation to a local one. The latter applying to special relativistic cases and the former to more "general" relativistic cases.

A:

The numbers $e^a_\mu(x^\nu)$ can indeed be understood as the components of a transformation matrix that makes a (coordinate-dependent) transformation from the coordinate basis on tangent space to the tetrad (in your case othonormal) basis. However, the issue with $e^a_\mu(x^\nu)$ is that it is simply not a matrix satisfying the properties of Lorentz-transform matrices. For a Lorentz-transform matrix $\Lambda^A_B$ we require 
$$\Lambda^A_B \Lambda^C_D \eta_{AC} = \eta_{BD}$$
But the tetrad components fulfill only $e^a_\mu e^b_\nu \eta_{ab} = g_{\mu\nu}$. This means, in particular, that we do not know how the matrices $J^a_\mu, K^a_\mu$ you posit above should look like.
One thing which turns out to be useful in several contexts, though, is the fact that the components of a transformation between two orthonormal tetrad bases $\Lambda^a_{\bar{a}} = e^a_\mu e^\mu_{\bar{a}}$ absolutely is a coordinate-dependent Lorentz-transformation matrix.

