Your desired final result does not seem to follow from the premises:
Clear[f, x, x1, x2, a];
f[x_] = 2 E^(2 x) - a (x - 1);
twoConds = f[x1] == 0 && f[x2] == 0;
thirdCond = a == (E^(2 x2) - E^(2 x1))/(x2 - x1);
counterExample = {x1 -> 2, x2 -> 1 - 1/2 ProductLog[-(2/E^2)],
a -> 2 E^4};
{twoConds, thirdCond} /. counterExample // Simplify