If $L(q,\dot{q},t)$ is a lagrangian of a system, then $L' = L + \frac{dF(q,t)}{dt}$ is also a valid lagrangian and both lagrangians will lead to the same equation of motion.
But, what if I choose $F(q,t)$ such that $L'=0$? \begin{equation} L + \frac{dF}{dt} = 0 \\ F = - \int{L dt} = -S \end{equation} where $S$ is the action.
Is this valid? If yes, can we not do this always? What does it mean?
The action $S$ is a functional of paths $q(t)$, not a function of the variables $(q,\dot{q},t)$, so the equality $F = -S$ makes no sense.
By claiming that you can choose $L = -\frac{\mathrm{d}F}{\mathrm{d}t}$, you already have restricted the original Lagrangian to be a total time derivative (of $-F$). This is not the case for Lagrangians that usefully describe physical systems, since it would mean that the action $$S[q(t)] = \int_{t_i}^{t_f}L(q(t),\dot{q}(t),t)\mathrm{d}t = - F(q(t_f),t_f) + F(q(t_i),t_i)$$ is only dependent on the start and end-points $(q(t_i),t_i)$ and $(q(t_f),t_f)$, so paths do not differ in their action and the principle of extremal action is useless.
If you can choose $F(q,t)$ such that $L'=0$, then $L$ can already be written in the form $$L\bigg(q(t),\dot q(t),t\bigg) = \frac{d}{dt}G\bigg(q(t),t)\bigg)$$ for some $G$. As a result, the action $$S[q] = \int_{t_1}^{t_2} L\bigg( q(t),\dot q(t), t\bigg)dt = G\bigg(q(t_2),t_2\bigg)-G\bigg(q(t_1),t_1\bigg)$$ is independent of the path $q$ (since the endpoints $q(t_2)$ and $q(t_1)$ are fixed by the boundary conditions). Since the action is the same for every path, there's no way to select a special one by demanding that $S$ be stationary with respect to small variations, so the action is not useful.
