Einstein's equations with $\Lambda=0$ are perfectly capable of describing an expanding spacetime. The cosmological principle leads to a metric of the form
$$ds^2 = -c^2 dt^2 + a(t)^2 d\Sigma^2$$
where $a(t)$ is the so-called scale factor which is conventionally set to $1$ at the present time, and $d\Sigma^2$ is a spatial 3-metric with constant curvature $\frac{k}{a^2} = \frac{1}{R_0^2}$ with $R_0$ the curvature radius. The application of Einstein's equations to this metric leads to the Friedmann equations, which govern the time-evolution of the scale factor:
$$\frac{\dot a^2 + kc^2}{a^2} = \frac{8\pi G\rho}{3} \qquad (1)$$
$$\frac{\ddot a}{a} = -\frac{4\pi G}{3}\left(\rho + \frac{3p}{c^2}\right)\qquad (2)$$
where $k\in\{0, \pm 1\}$ denotes the curvature of the spacelike hypersurfaces $\Sigma_t$, $\rho$ is the energy density (the $00$ component of the stress-energy tensor) and $p$ is the corresponding pressure. In order for the universe to be static, we would need to have that $\dot a =0$; equation (1) then implies that
$$a^2 = \frac{3kc^2}{8\pi G\rho} \implies \frac{1}{R_0^2} = \frac{k}{a^2} = \frac{8\pi G\rho}{3c^2}$$
and therefore that $k>0$, meaning that the universe is spatially closed - a sphere with radius $R_0$. However, this is not a steady-state configuration; if $\rho + \frac{3p}{c^2} \neq 0$, then $\ddot a < 0$ and the scale factor would begin to decrease. For ordinary (cold) matter and electromagnetic radiation, $p = 0$ and $p=\frac{\rho c^2}{3}$, respectively, so this would seem to be the case.
The addition of a cosmological constant $\Lambda$ solves this problem. The Friedmann equations become
$$\frac{\dot a^2 + kc^2}{a^2} = \frac{8\pi G\rho+\Lambda c^2}{3} \qquad (3)$$
$$\frac{\ddot a}{a} = -\frac{4\pi G}{3}\left(\rho + \frac{3p}{c^2}\right)+\frac{\Lambda c^2}{3}\qquad (4)$$
Choosing $\frac{\Lambda c^2}{3}=\frac{4\pi G}{3}\left(\rho + \frac{3p}{c^2}\right)$ makes $\ddot a=0$; setting $\dot a=0$ in equation (3) then yields
$$a^2=\frac{kc^2}{4\pi G\left(\rho + \frac{p}{c^2}\right)}$$
Assuming cold normal matter ($p=0$), this can be written
$$\frac{1}{R_0^2}=\frac{k}{a^2} = \frac{4\pi G \rho}{c^2} = \Lambda$$
This solution is brittle, however; note that if $a\rightarrow a+\delta a$, then $ \rho \rightarrow \rho + \delta \rho$ where
$$ \delta \rho = \frac{k}{a^2}\left(1-2\frac{\delta a}{a}\right)$$
which means via equation (4)
$$\ddot{\delta a} \propto \delta a$$
and so the equilibrium is unstable. Small perturbations will cause runaway expansion $(\delta a > 0 )$ or contraction $(\delta a < 0 )$.
The confirmation (Hubble, 1929) that the universe was not static - that the scale factor was indeed evolving with $\dot a > 0 $ - meant that this workaround using the seemingly arbitrary $\Lambda$ was unnecessary, and so Einstein abandoned it. It wasn't until 1998 that it was discovered that the expansion of the universe is accelerating, meaning that $\ddot a>0$.
This is a different beast. It either requires a cosmological constant which is sufficiently large to make the right hand side of equation (4) positive, or it requires a new kind of matter with equation of state $p < -\frac{1}{3}\rho c^2$ (or possibly some combination of the two). The basic form of the $\Lambda_{CDM}$ model considers only the cosmological constant; extensions or modifications of the model allow for different possibilities (see e.g. quintessence).
Ordinary matter and energy would always cause the Universe to contract and the rate of contraction would be increasing with time.
This is not true. If we assume a flat universe containing only cold baryonic matter (i.e. dust, with $p=0$), the Friedmann equations yield
$$a(t) \propto t^{2/3}$$
which increases forever without bound. Of course the real constitution of the universe is more interesting than this, but the point is that if $\dot a>0$ at some initial time, there is no cosmological constant needed to describe a universe which expands forever.
