Assuming you mean that $\mathscr X$ is an *algebraic* stack (e.g. in the sense of [Tag 026N]), then this is true, and follows relatively straightforwardly from the case of schemes.

Although you did not make a precise statement in the scheme case, it is indeed true that if $U \subseteq X$ is an open subscheme containing all *locally closed points*, then $U = X$. Moreover, locally closed points are the same thing as points that are (locally) of finite type [Tag 01TA].

Now let $\mathscr U \subseteq \mathscr X$ be an open substack, i.e. $j\colon\mathscr U \to \mathscr X$ is representable by open immersions. In particular, if $x \colon \operatorname{Spec} k \to \mathscr X$ is a point of finite type, the pullback along $\mathscr U \to \mathscr X$ gives an open immersion $U \hookrightarrow \operatorname{Spec} k$, and we say $\mathscr U$ *contains $x$* if $U \neq \varnothing$. If this holds for all finite type points, the question is whether $\mathscr U \to \mathscr X$ is an isomorphism.

By assumption, there exists a smooth surjection $f \colon X \to \mathscr X$ where $X$ is a scheme. Checking that a morphism is an isomorphism is smooth-local, so it suffices to check that the base change $U \to X$ of $j$ along $f$ is an isomorphism. Note that $U \to X$ is again an open immersion, so by the above it suffices to check that $U$ contains all points of finite type of $X$. If $x \colon \operatorname{Spec} k \to X$ is a point that is (locally) of finite type, then so is the composition $\operatorname{Spec} k \to X \to \mathscr X$, since $f$ is locally of finite type [Tags 06MH and 03YJ]. By assumption, the pullback of $U \to X$ along $x$ is an isomorphism, i.e. $x \in U$. $\square$