In practice, given a stress-energy tensor $T_{\mu\nu}$, we may attempt to find solutions to the Einstein field equations using perturbation theory. The basic idea is to expand around a known solution $g_{\mu\nu}$ by a perturbation $h_{\mu\nu}$. In the case of a flat background,
$$\delta G_{\mu\nu} = 8\pi G \delta T_{\mu\nu} = \partial_\mu \partial_\nu h - \partial_\mu \partial_\alpha h^\alpha_\nu -\partial_\nu \partial_\alpha h^\alpha_\mu + \partial_\alpha \partial^\alpha h_{\mu\nu} - \eta_{\mu\nu} \partial_\alpha \partial^\alpha h + \eta_{\mu\nu} \partial_\alpha \partial_\beta h^{\alpha \beta}.$$
In some cases, one may solve the equations exactly, or more typically, employ numerical methods. Another alternative approach when faced with a stress-energy tensor is to try to determine the symmetries the metric may have, and then plug in an ansatz for the metric to yield a set of differential equations which may be more tractable, analytically and numerically.
Yet another alternative is to generate new solutions from old ones through the introduction of pseudopotentials, a method due to Harrison, Eastbrook and Wahlquist, called the method of prolongation structures.
There are several other methods such as those that rely on Lie point symmetries of differential equations. There is also a Backlund transformation, which relies on identifying a simpler differential equation whose solution satisfies a condition involving the solution to the harder problem.
These methods are too involved to present here and require a significant background. They are explained in Exact Solutions to the Einstein Field Equations by H. Stephani et al.
Addressing your other question, if given a metric $g_{\mu\nu}$, of course one can compute $T_{\mu\nu}$ through the Einstein field equations, you just plug it in and tediously compute all the curvature tensors. There are strictly speaking probably some requirements on the functions in $g_{\mu\nu}$, but you can get away with most things, even distributions, such as a delta function, which may lead to a stress-energy tensor describing a brane.
Best Answer
The metric tensor is unitless. That can be seen from the fact that $g_{\mu\nu}v^\mu v^\nu$ gives the square of the four-vector length of $v$, and thus has the unit of $v^2$.
The scalar curvature is a contraction of the Ricci tensor. A contraction doesn't change the units. Also the Ricci tensor is a contraction of the Riemann tensor.
The Riemann tensor is made of coordinate derivatives of the connection coefficients, which are made of coordinate derivatives of the metric. Since each coordinate derivative adds a unit $m^{-1}$, the Ricci tensor and curvature scalar have both unit $\mathrm{m}^{-2}$.
The cosmological constant then of course also has to have the unit $\mathrm m^{-2}$, so that the units match.
$T$, the stress-energy tensor, has the unit of energy density, or pressure (both are actually the same unit, if you look closer), that is, $\mathrm J/\mathrm m^3$ or $\mathrm N/\mathrm m^2$.