Does almost constant energy density only require that $ \omega \approx-1 $? 
*

*From continuity equation 


\begin{equation}
\dot{\rho}=-3H(p+\rho)
\end{equation}
$ p+\rho\approx0 $ will give $ \dot{\rho} \approx 0 $ which means almost constant energy density 


*

*From equation of state 


$$ \omega = \frac{p}{\rho} $$
$ \frac{p}{\rho}\approx-1 $ will give $\omega \approx-1$ which means almost constant energy density 

I’d use a scalar field as an example which is defined as$$
\rho=\frac{1}{2}\dot{\phi}^2+V(\phi)~,~~p=\frac{1}{2}\dot{\phi}^2-V(\phi)~,
$$ 
But suppose that $~ \rho=10 + 10^6$,$~~p=10- 10^6$ . We can see that $ \omega=\frac{p}{\rho}\approx-1 ~$ but not $~p+\rho\approx0 $.
Does almost constant energy density only require that $ \omega \approx-1 $ without having to consider $ p+\rho\approx0 $ ?
 A: You are missing units! Whenever you say that some quantity is big or small, you need to say relative to what. The only quantities that have an "absolute" size are dimensionless quantities.
Say that actually $\rho = (10 + 10^6)\, \mathrm{eV^4}$ and $p = (10-10^6)\, \mathrm{eV^4}$, in natural units. Then since $\omega$ is dimensionless we can indeed say that $\omega \approx -1$, or, more precisely, that $|\omega+1| \ll 1$. But $p+\rho = 20\, \mathrm{eV^4}$, and this quantity by itself is neither big nor small; you need to compare it to something else. In this case, the correct statement is that $|p+\rho| \ll \rho$ or $|p+\rho| \ll |p|$, since $p$ and $\rho$ are the dimensionful quantities that provide a scale.
Edit: A couple of comments. You say that $20\,\mathrm{eV^4}$ is not approximately zero, and I see why you say that. But converting to SI units, $20\,\mathrm{eV^4}$ is around $4\times 10^{-15}\, \mathrm{kg/m^3}$. Now it seems pretty small, doesn't it? The moral is that we can't just say that physical quantities with units are big or small, because the numerical value depends on what units we use. We have to compare with something else.
It doesn't make sense to just ask that $\dot\rho$ be small. No matter how slow $\rho$ changes, if we wait a long enough time it can change a lot. What we need to do is identify our time scale $T$: something that gives us a rough idea of just how much time we are willing to wait. Here it is given by $T = 1/H$, the inverse of the Hubble parameter, and what we need is that over time intervals not much longer than $T$, $\rho$ doesn't change much compared to itself. That is, $\Delta \rho \ll \rho$, with $\Delta \rho \approx \dot{\rho} T = \dot{\rho}/H$. So we see that the condition we want should properly be written $|\dot{\rho}/H| \ll \rho$, or, using the differential equation, $|\rho+p| \ll \rho$, like I said before.
