The result that you are looking for is not in Élie Cartan's 1936 book La topologie des groupes de Lie because it was not known to be true at the time the book was written. Indeed, as Cartan remarks in the book (which is the lecture notes from an October 1935 conference where he spoke), it was not even known at that time that every Lie algebra over the reals was the Lie algebra of a Lie group. At least, it was not known by him.
However, I. D. Ado, then a student at Kazan State University, had, in 1935, published (in Russian) a proof of his now famous result that every Lie algebra over the reals has a faithful, finite dimensional representation, which implies the existence of a (unique up to isomorphism) connected and simply-connected Lie group with any given Lie algebra. (Apparently, Cartan was not aware of Ado's work when his book went to press, but he was certainly aware of it sometime before 1938; it's hard to say exactly when.)
Assuming the existence of a connected, simply-connected Lie group $G$ whose left-invariant vector fields have a basis $Y_i$ satisfying $[Y_i,Y_j]= c^k_{ij} Y_k$ (I omit the summation sign), one can sketch the proof of the result you want (assuming that $M$ itself is connected) as follows:
On the product manifold $P = M\times G$ consider the vector fields $Z_i = X_i + Y_i$ (where I am identifying the tangent space $P_{(m,g)}$ with $T_mM\oplus T_gG$ in the obvious way), these $n$ linearly independent vector fields satisfy $[Z_i,Z_j]= c^k_{ij} Z_k$ and hence span an $n$-plane field that is integrable. Thus, by the Frobenius theorem, $P$ is foliated by the $n$-dimensional 'leaves' that are everywhere tangent to the $Z_i$. Fix an $m\in M$ and let $L\subset M\times G$ be the leaf of this foliation that passes through $(m,e)$.
The projection of $L$ to $G$ is a local diffeomorphism; we want to show that it is a covering map, i.e., that $\pi:L\to G$ has the homotopy lifting property.
Here is where the compactness of $M$ comes in. Since $M$ is compact, the vector fields $X_i$ are complete (i.e., their flows exist for all time). Using this, it is easy to show any differentiable curve $\gamma:[0,1]\to G$ such that $\gamma(0) = \pi(p)$ for $p\in P$ can be lifted uniquely to a curve $\tilde\gamma:[0,1]\to P$ such that $\tilde\gamma(0)=p$ and $\pi_2\circ\tilde\gamma = \gamma$ and $\tilde\gamma$ is everywhere tangent to the $n$-plane field spanned by the $Z_i$. (In fact, you just write $\gamma'(t) = a^i(t) Y_i(\gamma(t))$ for some functions $a^i$ and then let $\tilde\gamma$ satisfy $\tilde\gamma(0)=p$ and $\tilde\gamma'(t) = a^i(t) Z_i(\tilde\gamma(t))$.) The fact that $L$ is a covering map follows immediately.
Since $G$ is connected and simply connected, $\pi$ must be a diffeomorphism, and, hence, the graph of a smooth mapping $f:G\to M$ such that $X_i$ is $f$-related to $Y_i$. Now, we can use the same trick as above to show that for any smooth curve $\gamma:[0,1]\to M$ with $\gamma(0) = f(g)$, there is a smooth curve $\tilde\gamma:[0,1]\to G$ such that $\tilde\gamma(0)=g$ and $f\circ\tilde\gamma = \gamma$. Since $M$ is connected, every point can be connected to $f(e)$ by a smooth curve, so $f$ is surjective and is a covering map.
Finally, let $H = f^{-1}(m)\subset G$. Then $H$ is discrete because $f$ is a covering map. Moreover, if $h\in H$ and $\lambda_h:G\to G$ is left-multiplication by $h$, then $f\circ \lambda_h(e) = f(h) = m$ and, since the vector fields $Y_i$ are left-invariant, it easily follows that the graph of $f\circ\lambda_h$ in $P$ is a leaf of the $n$-plane field spanned by the $Z_i$ that contains $(m,e)$, so it must be equal to $L$. Consequently, $\lambda_h(H) = H$ (implying that $H$ is a subgroup of $G$) and $\lambda_h$ is a deck transformation for the covering map $f:G\to M$. Thus, $H$ is a discrete subgroup of $G$, and $M$ is diffeomorphic to $H\backslash G$.
You might want to think about whether you can extend this result to the case that $M$ is not connected (in which case, of course, $G$ would have to be disconnected). Consider the case where $M$ is the disjoint union of two circles and the flow of $X_1$ has period 1 on one of the circles and period $\pi$ on the other.
Now, on to your second question, which is more subtle. We know that not every compact, connected, orientable $3$-manifold $M$ is homogeneous, so you can't hope to find a frame field $X_i$ for which the $\alpha^i_{jk}$ are constants in that generality. You might hope to impose some weaker condition $C$ on the $\alpha^i_{jk}$ so that you could guarantee that a frame field satisfying $C$ would always exist and yet that $C$ would be 'geometric' enough that you could use it to recognize pieces into which $M$ could be cut as 'geometric'. Or you might hope to find some expression in the $\alpha^i_{jk}$ that you could integrate over $M$ (using the volume form for which the $X_i$ are unimodular) to give you some measure of the 'complexity' of the frame field $X_i$ that you could try to minimize by some kind of gradient flow. If the minima of this functional were well-enough behaved, maybe that would give you a clue as to how to use such a minimizer to cut $M$ into geometric pieces. (Or you could follow the strategy of Perelman's argument and try to understand the singularities that develop when you attempt to flow to a minimizer.)
You could even give yourself a 'headstart' by imposing some vanishing at the start. For example, it's not hard to show that you can always choose a frame field $X_i$ so that $\sum_i\alpha^i_{ij} =0$ for all $j$, which gets rid of $3$ of the $9$ components right off the bat.
The main problem, though, will be finding conditions on the $\alpha$ that are somehow attainable or approachable and yet give you geometric information. The nice thing about metrics, as in Hamilton's and Perelman's approach(es), is that we know a lot about Riemannian geometry. We know relatively little about how to interpret the invariants of frame fields geometrically except in very special circumstances (such as the $\alpha$s being constant), and they are (so far) just too special to be of much use in tackling something like the Poincaré Conjecture or the Geometrization Theorem.
I'm not saying that 'frame field geometry' won't get you anywhere in studying the topology of $3$-manifolds, I just think that the necessary work of developing the foundations of such a geometric theory hasn't been sufficiently pursued that we can have a good sense of what specific things to try. (And you can see, from what I wrote above, that there are many possible things to try.)
