Perhaps an example will help? Consider the momentum operator $A=-id/dx$, the eigenstates $|a\rangle$ are plane waves $\langle x|a\rangle=(2\pi)^{-1/2} e^{iax}$, the expansion (1) of an arbitrary state $|\psi\rangle$ in the $|a\rangle$ basis is the Fourier integral $$\langle x|\psi\rangle=\int_{-\infty}^\infty da\,\psi(a)\langle x|a\rangle=(2\pi)^{-1/2}\int_{-\infty}^\infty da\,e^{iax}\psi(a),$$ and the completeness relation reads $$\int_{-\infty}^\infty da\,\langle x|a\rangle\langle a|x'\rangle=(2\pi)^{-1}\int_{-\infty}^\infty da\,e^{ia(x-x')}=\delta(x-x')=\langle x|I|x'\rangle,$$ with $\delta(x-x')$ the Dirac delta function distribution.

So yes, you need to equip the Hilbert space with a distribution theory, so a rigged Hilbert space is needed (as for any unbounded observable with a continuous spectrum).

Perhaps an example will help? Consider the momentum operator $A=-id/dx$, the eigenstates $|a\rangle$ are plane waves $\langle x|a\rangle=(2\pi)^{-1/2} e^{iax}$, the expansion (1) of an arbitrary state $|\psi\rangle$ in the $|a\rangle$ basis is the Fourier integral $$\langle x|\psi\rangle=\int_{-\infty}^\infty da\,\psi(a)\langle x|a\rangle=(2\pi)^{-1/2}\int_{-\infty}^\infty da\,e^{iax}\psi(a),$$ and the completeness relation reads $$\int_{-\infty}^\infty da\,\langle x|a\rangle\langle a|x'\rangle=(2\pi)^{-1}\int_{-\infty}^\infty da\,e^{ia(x-x')}=\delta(x-x')=\langle x|I|x'\rangle,$$ with $\delta(x-x')$ the Dirac delta function distribution.

So yes, you need to equip the Hilbert space with a distribution theory --- as for any unbounded observable with a continuous spectrum a rigged Hilbert space is needed.

I found this discussion of the Dirac bra-ket formalism in a continuous spectrum instructive.