“Why does igniting propellant provide more thrust?”
It doesn't!
At least with the specified parameters. Consider the following control volume:
For a perfectly expanded case with choked flow, the momentum thrust $\dot{m}V_e$ will be equal to the pressure thrust $F$. Where pressure thrust is defined as:$$F = \int P_xdA$$
where $P_x$ is the axial component of static pressure acting on a given piece of thrust chamber wall $dA$.
Combusting propellants increases the total temperature of a fluid, so you are asking about two cases: one with a very high $T_o$ and one with a low $T_o$ and both with the same total pressure $P_o$. Lets investigate the pressure thrust first.
To calculate pressure thrust, we need to know the static pressure at every axial location in the thrust chamber. For isentropic flow, the static pressure is related to the Mach number by
$$P = P_o\left(1+\frac{\gamma - 1}{2}M^2\right)^{\frac{-\gamma}{\gamma-1}}$$
and the Mach number is related to the area ratio at a given axial location by
$$\frac{A}{A^*} = \left(\frac{\gamma+1}{2}\right)^{-\frac{\gamma+1}{2(\gamma-1)}}\frac{\left(1+\frac{\gamma-1}{2}M^2\right)^{\frac{\gamma+1}{2(\gamma-1)}}}{M}$$
A nasty equation. Here $\gamma$ is the ratio of specific heats and $\frac{A}{A^*}$ is the ratio of channel area to throat area. What can we glean from these equations? Mach number is purely a function of the area ratio and of gamma, so for the hot and cold case, the Mach distribution along the thrust chamber will be the same. The static pressure is a function of total pressure, gamma, and Mach, none of which change between the two cases, so the static pressure distribution will also be the same! (neglecting the weak dependence of gamma on temperature).
If the pressure distribution is the same, then the thrust must also be the same between the hot and cold cases.
Now, to the source of your confusion: what about the momentum thrust!? Surely this increases! Lets check.
As you mentioned, the exit velocity for the hot case will increase according to
$$V_e = M_e\sqrt{\gamma RT_e}$$
where the subscript $e$ denotes conditions at the exit plane and $T_e$ is the exit static temperature. This follows a similar relationship as the static pressure with
$$T = T_o\left(1+\frac{\gamma-1}{2}M^2\right)^{-1}$$
Clearly the exit static temperature is directly proportional to the total temperature, so the hot case will have a higher gas exit velocity. So you say "Yes, that is why the thrust must be higher!", but calm your pants; what about the mass flow?
The mass flow for a choked nozzle is defined by
$$\dot{m} = \frac{A^*P_o}{\sqrt{T_o}}\sqrt{\frac{\gamma}{R}}\left(\frac{\gamma+1}{2}\right)^{-\frac{\gamma+1}{2(\gamma-1)}}$$
where $R$ is the ideal gas constant. Aha! Perhaps things become clearer: the mass flow is inversely related to the total temperature. Could the decrease in mass flow exactly counter the increase in velocity? Solving for the momentum thrust, $\dot{m}V_e$ by crudely combining these equations gives
$$\dot{m}V_e = A^*P_oM_e\sqrt{\frac{\gamma^2RT_o}{RT_o\left(1+\frac{\gamma-1}{2}M_e^2\right)}}$$
Well well well, would you look at that: the term $RT_o$ cancels from the momentum thrust. The momentum thrust will also be the same for the two cases!
Whew, crisis of aerothermal physics averted. But wait, if thrust is the same, why does anybody bother with those extremely high temperatures? The answer is mass flow efficiency. As we discovered, the higher temperature case used lower mass flow for the same thrust. This gives a higher specific impulse and makes a rocket using this engine much lighter and cheaper than for the same journey using cold gas.
