The answer to all questions is yes, and in fact a non-constant dark energy only makes sense if it is interpreted as part of the energy-stress tensor. The Einstein Field Equations for the FLRW metric yield the continuity equation
$$c^2\frac{\text{d}(\rho a^3)}{\text{d}t} + p\frac{\text{d}(a^3)}{\text{d}t} = 0,$$
which can also be written as,
$$
\dot{\rho} + 3\frac{\dot{a}}{a}\left(\rho + \frac{p}{c^2}\right)=0.
$$
Here, $\rho$ is the density, $p$ is the pressure, and $a$ is the scale factor. This equation can be derived from the Friedmann equations or from the conservation of the stress-energy tensor (see this post). We can assume that, for most of the history of the universe, the various constituents (radiation, matter, and dark energy) behave independently, so that the continuity equation holds for each constituent separately.
Next, we need to postulate an equation of state between $p$ and $\rho$. One usually assumes a linear relation of the form $p = w\rho c^2$, where in general $w$ is a function of the scale factor (or equivalently, of redshift). The continuity equation then becomes
$$
\frac{\text{d}\rho}{\rho} = -3\big[1+w(a)\big]\frac{\text{d}a}{a},
$$
with solution
$$
\rho(a) = \rho_0 \,\exp\left(-3\int_1^a\frac{1+w(a)}{a}\text{d} a\right),
$$
where $\rho_0$ is the present-day value. Note that $w(a)\equiv 1/3$ and $w(a)\equiv 0$ yield the familiar densities of radiation and matter, respectively:
$$
\rho_r(a) = \rho_{r,0}\,a^{-4},\qquad \rho_m(a) = \rho_{m,0}\,a^{-3}.
$$
The value $w(a)\equiv -1$ corresponds with a cosmological constant:
$\rho_\Lambda(a) \equiv \rho_{\Lambda,0}$. The simplest and most common non-constant dark energy model is of the form
$$w(a) = w_0 + (1-a)w_a,$$
for which
$$
\rho(a) = \rho_0 \,a^{-3(1+w_0+w_a)} e^{3(a-1)w_a}.
$$
Other common models are listed in this article.