@Buzz Bloom PeterDonis is right that these aren't two different concepts, they are just two different ways of expressing the same concept. I don't know if what I'm about to say fully answers your question. But the relationship between the cosmological constant and the dark energy density can be derived from the First Friedmann Equation. This equation describes the relationship between the dynamics of the Universe's expansion, and the density of its mass-energy constituents. In a flat Universe, it is given by:
$$H^2 = \left(\frac{\dot{a}}{a}\right)^2 = \frac{8\pi G}{3}\rho_\mathrm{tot}$$
Here, ##a## is the scale factor, and ##\rho_\mathrm{tot}## is the total energy density of
all the constituents making up the "perfect" fluid considered to permeate the Universe. (I think it should actually be ##\rho c^2##, but I'm used to using ##c=1## units, and not caring about the distinction between mass density and energy density). So anyway, since I have no curvature term in there (i.e. I'm assuming a flat Universe), it follows also that the total density is equal to the critical density i.e. ##\rho_\mathrm{tot} = \rho_\mathrm{crit}##, or ##\Omega_\mathrm{tot} = 1##. Substituting in the critical density on the righthand side, we can also derive that, at the present day (where ##H = H_0##):
$$\frac{3H_0^2}{8\pi G} = \rho_\mathrm{crit}$$
Let's file that result away for later.
Going back to the First Friedmann equation, we can break up the total density into individual constituents like matter (both baryonic & dark), radiation (photons & relativistic particles), and dark energy: ##\rho_\mathrm{tot} = \rho_m + \rho_r + \rho_{de}##. So the Friedmann equation becomes:
$$H^2 = \frac{8\pi G}{3}\left(\rho_m + \rho_r\right)+ \frac{8\pi G}{3}\rho_{de}$$
Now, compare this in your head to a version of the Friedmann equation that only considered a Universe with matter and radiation, but then had the cosmological constant term of ##\Lambda/3## tacked on to the righthand side as a sort of "fudge factor". (I think it's actually ##\Lambda c^2/3##. But again, factors of ##c^2## will float around as they do. I care not, for I have set them to unity). The idea here is that if the observed data showing accelerating expansion, which strongly favour non-zero Lambda, can be attributed to some actual substance or field (dubbed "dark energy") permeating space with constant density, then it follows just by comparison of the dark energy term with the Lambda one that:
$$\frac{8\pi G}{3}\rho_{de} = \frac{\Lambda}{3}$$
therefore:
$$\Lambda = 8\pi G \rho_{de}$$
So you can see that ##\Lambda## and ##\rho_{de}## (what you call ##\rho_{\Lambda}##) are the same thing, up to a constant of proportionality. And they are both dimensionful constants, it's just that ##\Lambda## absorbs the extra constants of proportionality (##8\pi G##) into the density, giving the cosmo-constant term the same dimensions as the lefthand side: dimensions of "Hubble parameter squared" i.e. 1/time².
We can also relate ##\Lambda## to the dimensionless
density parameter for dark energy just by applying the
definition of the density parameter for a constituent. It's the ratio of the density of that constituent to the critical density. Here's where we use that result that we filed away. At the present day, this would be given by:
$$\Omega_{\Lambda} \equiv \frac{\rho_{\Lambda}}{\rho_\mathrm{crit}} = \left(\frac{\Lambda}{8\pi G}\right)\left(\frac{8\pi G}{3H_0^2}\right) = \frac{\Lambda}{3H_0^2}$$
This expression for ##\Omega_{\Lambda}## is the same as the one in your OP (up to some factors of ##c## which, again, I don't care about here). And, as we established in my discussion above, ##H_0^2## and ##\Lambda## have the same dimensions, so we know that the units cancel, leaving ##\Omega_{\Lambda}## dimensionless as expected.
Hope that helps, somewhat!
EDIT: note that ##\rho_{de}## may not actually be constant with time. We don't know for sure yet. We need more and better data to be sure, e.g. from cosmology survey telescopes that map out large-scale structure, like WFIRST/Roman and Euclid. If ##\rho_{de}## is constant, then it corresponds to the "Lambda" (cosmo-constant) term in the Friedmann equations. This is the simplest model. But if ##\rho_{de}## varies with time, then it enters into the equation in a scale-factor-dependent way.
