The electron configuration of the hydrogens are approximately
$$\ce{\underset{1s}{[\uparrow]}}$$
although some of the charge is pulled off into the oxygen.
The electron configuration of a neutral oxygen atom is
$$[\ce{He}] \underset{2s}{[\uparrow \downarrow]} \underset{2p}{[\uparrow \downarrow \vert \uparrow \vert \uparrow]}$$
In $\ce{H2O}$ each free electron in a hydrogen pairs up with a free electron in an oxygen (approximately.) So how can a $\ce{H3O^+}$ ion form?
In $\ce{H3O^+}$ there is a missing electron so the oxygen could have an electron configuration of
$$[\ce{He}] \ce{\underset{2s}{[\uparrow \downarrow]}} \underset{2p}{[\uparrow \vert \uparrow \vert \uparrow]}$$
which allows for a hydronium ion to form.
Now in an actual hydronium ion the oxygen would have hybridized orbitals of the form
$$[\ce{He}] \underset{(2s)(2p)^3}{[\uparrow \downarrow \vert \uparrow \vert \uparrow \vert \uparrow]}$$
which is a tetrahedral arrangement.
Now this is only an approximation. A more in depth look would go deeper into molecular orbital theory and bond symmetries but that gets annoying for polyatomic molecules.