why do we look at dark energy as an energy?
Since relativistic physics almost always takes units of $c = 1$, there is no difference between units of energy density and mass density. Conceptually, however, the gravitational charge is energy, not mass.

In general relativity, a perfect fluid has no viscuous shear stresses and no heat flux, and can be described by the stress-energy tensor $T^{\mu\nu} = (\rho + p)u^\mu u^\nu + pg^{\mu\nu}$, where $\rho$ is density, $p$ is pressure, and $u^\mu$ is the fluid four-velocity. In the case of vanishing pressure, the result is called a dust, and is appropriate for modeling cold dark matter, galaxies on the cosmological scale, etc. On the other hand, a radiation fluid has $\rho = 3p$, and is appropriate for modeling incoherent electromagnetic radiation, or a fluid made of strongly relativistic particles (e.g., hot dark matter), and so forth.

Thus, in the specific context of cosmology, the matter-dominated regime has $p\sim 0$ and radiation-dominated regime has $p\sim\rho/3$. Note that in the latter case, the fluid particles have zero (or at least negligible) mass. In-between, we can talk about a fluid of particles that are not strongly relativistic, and hence have some mass.

On the other hand, the standard model of dark energy is a cosmological constant, which is equivalent to a perfect fluid $p = -\rho$. It does not fit into the above scale; it is not matter in the above sense, nor is radiation, nor anything in-between. Furthermore, it has the peculiar property of being exactly the same in every local inertial frame. This frame-independence enables us to think of the cosmological constant as an intrinsic energy density of spacetime, but it also makes it inappropriate to think of it as a collection of fluid particles that have some rest frame.

For your purposes, this important because $E = mc^2$ is only valid in a rest frame.