Interesting idea! Let's check the numbers.
For this to work at Earth's surface, the electric force has to cancel $P = 10^5 N/m^2$.
The electric field $E = \sigma / \epsilon_0$ in the case (no field inside, all the field goes outside).
For an area $A$, the force on a conductor due to a field $E$ is $F = QE/2 = \sigma A E / 2$ so the force per unit area is
$P' = F/A = \sigma E / 2 = \epsilon_0 E^2$
where I've expressed it in terms of $E$ so we can solve for the electric field $E$ needed to set the electric "pressure" equal to atmospheric pressure (ideally, you'd want a bit more to keep things stable):
$ \epsilon_0 E^2 = 10^5$
$ E^2 = 10^5 / 9\times10^{-12} \sim 10^{16}$
So the electric field you need is about $10^8$ volts/meter. That's a lot. It's 3,000 times more than the 30kV/m breakdown voltage of air.
What does that mean? It means that the charge on your balloon is going to instantly spark off into the air and the balloon won't be held up against the atmosphere any more.
Note that insulating the balloon won't help. The electric field has to go off to infinity to hold up the balloon against pressure. If you add insulation against sparking, several things can happen:
That insulation will still have the electric field outside it, so where the insulation ends, the sparking starts, and the balloon ends up neutralized.
Or you make the insulation so thick, without adding weight, that the surface area of the ballon is much, much, much (3000X) times bigger than the balloon itself, spreading the field out and reducing it.
So it looks like this isn't going to work at Earth's surface. Bottom line, the atmosphere pushes too hard, electrostatic forces are too weak, and the atmosphere isn't a good enough insulator.
But it's still a cool idea.