Vapour Pressure of A Solution
This is where your logic fails:

since the concentration of reactants and products remains the same since B is different from A

As soon as you mix two components you change the concentration of both, you effectively dilute all components.
Otherwise your intuition is correct. The gases can often be assumed to act independently. This is true if the solution is ideal (that is, assuming interactions between all molecules in solution are roughly the same). Then each component equilibrates independently with its vapor. However, the vapor pressure of each component is in general not the same as it would be if it were pure. Instead, for an ideal solution the partial pressure of each component follows Raoult's law $p_i=p^*_i\chi_i$, where $p^*_i$ is the partial pressure of the pure component and $\chi_i$ is the mole fraction in solution.
The total pressure above the solution is then the sum of the partial pressures of the components (following Dalton's law, where $\chi_i^g$ is the mole fraction in the gas phase):
$$ \begin{align} p &amp;= \sum_i p_i = \sum_i \chi_i^g p \\ &amp;= \sum_i \chi_i p^*_i \end{align} $$