I would like to know what are all the real three dimensional Lie groups (simply connected) that can act transitively and locally freely on a compact three dimensional manifold?
This is equivalent to finding all the real three dimensional simply connected Lie groups $G$ having a discrete subgroup $\Gamma$ such that $G/\Gamma$ is compact?
Some known examples are: $\mathbb{R}^3$, $\mathbb{S}^3$, $\sf{SL}(2,\mathbb{R})$, and the Heisenberg group.
A connected Lie group with a lattice is unimodular. In dimension $\le 3$, we can list them and check that they admit cocompact lattices.
The full list:
1) compact ones ($\{1\}$ is a cocompact lattice): tori of dimension $0,1,2,3$, $\mathrm{SO}(3)$ and its two-fold covering $\mathrm{SU}(2)$.
2) non-compact non-solvable ones: connected coverings of $\mathrm{PSL}_2(\mathbf{R})$. The inverse image of a surface group is a cocompact lattice.
3) simply connected solvable ones:
beyond $\mathbf{R}^i$ for $i=0,1,2$, these are all of the form $\mathbf{R}^2\rtimes\mathbf{R}$. To be unimodular, the action has to be of determinant 1. This gives 4 cases:
3a) trivial action $\mathbf{R}^3$: $\mathbf{Z}^3$ is a cocompact lattice;
3b) unipotent action: Heisenberg group. The integral Heisenberg group is a cocompact lattice
3c) rotation action (hence not faithful): some central subgroup $\mathbf{Z}^3$ is a lattice.
3d) diagonal determinant 1 action: this is called group SOL. Each $\mathbf{Z}^2\rtimes\mathbf{Z}$ is a cocompact lattice therein, when the action is by powers of a matrix in $\mathrm{SL}_2(\mathbf{Z})$ with trace $\ge 3$.
4) The non-simply connected solvable non-compact ones are covered by the previous ones (modding out by a discrete central subgroup).
4a) Covered by 3a: this gives $\mathbf{R}^3/\mathbf{Z}$ and $\mathbf{R}^3/\mathbf{Z}^2$, which admit $\mathbf{Z}^2$ and $\mathbf{Z}$ as cocompact lattices.
4b) Covered by 3b: this gives Heisenberg modulo central $\mathbf{Z}$, which admits $\mathbf{Z}^2$ as cocompact lattice.
4c) Covered by 3c: this gives the oriented isometry group of the plane $\mathrm{SO}(2)\ltimes\mathbf{R}^2$ and its finite cover, which admit $\mathbf{Z}^2$ as cocompact lattice.
Remark: In dimension 4, there are continua of non-isomorphic unimodular groups of the form $\mathbf{R}\ltimes\mathbf{R}^3$, diagonal action. Most of these have no lattices.
$\DeclareMathOperator\PSL{PSL}$Note that since you ask for simply-connected groups, you should use the universal cover $\widetilde\PSL(2, R)$ of $\PSL(2, R)$ rather than $\operatorname{SL}(2, R)$.
The question of what compact $3$-manifolds are quotients of a simply-connected Lie group by a discrete subgroup was answered by F. Raymond and A. Vasquez in "3-Manifolds whose universal coverings are Lie groups" published in Topology and its Applications 12 (1981) 161-179. (In fact they implicitly answered the more general question in which $G/\Gamma$ has finite volume.) Almost all of the resulting three manifolds are Seifert manifolds and $\Gamma$ is described in terms of the Seifert invariants of the quotient. The few cases not describable this way are quotients of solvable Lie groups. A Lie group that admits a finite volume quotient by a discrete group is unimodular (this is proved in Milnor's paper Curvatures of left invariant metrics on Lie groups) and the unimodular three-dimensional Lie groups were classified by Bianchi (L. Bianchi, Sugli spazi a tre dimensioni che ammettono un gruppo continuo di movimenti, Memorie della Società Italiana delle Scienze (detta dei XL) 11 (1898), 267–352). In addition to the abelian case, nilpotent case (Heisenberg), and the semisimple cases $\operatorname{SU}(2)$ and $\widetilde\PSL(2, R)$, $G$ can also be the (universal cover of the) groups of Euclidean motions of the plane or the group of affine motions preserving a split signature bilinear form on the plane (these are solvable but not nilpotent).
