You lost the [itex]\delta[/itex].

[tex]\delta S' = \delta S + \delta F(q,\dot{q})\bigg |_{t_1}^{t_2}[/tex]

And [itex]\delta F[/itex] isn't necessarily trivial under variations of trajectory. At very least, [itex]\dot{q}(t_2)[/itex] is not fixed by boundary conditions.

But anyways, the statement "Lagrangian is invariant under translation," means that if you shift the Lagrangian itself

*and* the solution, the shifted solution is a solution to shifted Lagrangian. In other words, yes, equations of motion remain the same.

This is all made far more explicit if you look at where it comes from, namely,

Noether's Theorem.

Edit: Scratch that. I should be more careful when I answer questions this late. You don't shift the Lagrangian. Just the variables. For example, if you have two masses on a spring, the potential energy depends only on difference of positions. So if you shift both coordinate variables by same amount, the total Lagrangian remains exactly the same. In contrast, if you have an mgy term for gravitational potential, shifting y results in change in Lagrangian. But that makes sense. Vertical momentum is not conserved, because we are ignoring momentum transfer to Earth when we write potential this way.