In special relativity, the spacetime interval between the same two events measured by two different inertial observers are numerically the same: $$c^2dt^2-|d\vec{x}|^2=c^2dt^{'2}-|\vec{dx'}|^{2}.$$
In general relativity, does the same thing remain true i.e. if two arbitrary observers measure the same two events, does their spacetime intervals $g_{ab}(t,|\vec{x}|)dx^a dx^b$ and $g_{ab}(t',|\vec{x}'|)dx^{'a} dx^{'b}$ are also numerically equal?
Yes they are the same, but just to be clear, you need to indicate in the notation that the second set of metric components are not the same as the first set of metric components. This can be done by using a prime, as in $$ g_{ab} dx^a dx^b = g'_{ab} dx'^a dx'^b $$ This equality is one of the foundational ideas of General Relativity. It announces that we are dealing with a Riemannian or pseudo-Riemannian manifold. In tensor analysis this expression has the correct form to give an invariant scalar.
