Does the van der Waals equation remain valid when repulsive intermolecular forces dominate?
You need to consider the $b$ term, as well as $a$.

Rearranging the van der Waals equation to solve for $p$ gives

$$p = \frac{RT}{V_\mathrm{m} - b} - \frac{a}{V_\mathrm{m}^2}$$

You are right that a positive $a$ will always decrease the pressure, compared to the ideal gas result.

But a positive $b$ will act to increase the pressure compared to the ideal gas result.

At high pressures (small molar volumes), the effect from finite molecular volume ($b$) outweighs the effect of intermolecular attraction ($a$).