Convenient factorisation of position basis of the Hilbert space of lattice-periodic system

I am dealing with a one-dimensional system that has a discrete translational symmetry, i.e. the Hamiltonian has the form
$$
hat H= frac{hat p^2}{2m}+V(x)
$$
where $V(x+a)=V(x)$ and $hat H$ is an operator acting on the Hilbert space $mathcal{H}_T$.

Expressing the identity operator in the position basis one has:
$$
hat {1}_{mathcal{H}_T}=int_{mathbb{R}} big|xbig>big<xbig| dx tag{1}
$$

Now, an equivalent way to express Eq. (1) is
$$
{1}_{mathcal{H}_T}=sum_{ninmathbb{Z}} int_{0}^a big|ncdot a+ybig>big<ncdot a+ybig| dy tag{2}
$$

Intuitively, I would expect to be possible to express the bra/ket appearing in the identity operator (2) as:

$$big|ncdot a+ybig>_{mathcal{H}_T}=big|ncdot abig>_{mathcal{H}_d}otimes big|ybig>_{mathcal{H}_c}tag{3}$$
so that the total Hilbert space can be obtained as
$mathcal{H}_T=mathcal{H}_dotimes mathcal{H}_c$, i.e. as
a tensor product of an Hilbert space having discrete states ($big|ncdot abig>inmathcal{H}_d$ with $nin mathbb{Z}$) and an Hilbert space in which the position operator has bounded eigenvalues, ($big|ybig>inmathcal{H}_c$ with $yin [0,a)$)

In formulas, I am asking if it is possible to factorise the operator (2) as follows
$$
{1}_{mathcal{H}_T}=sum_{ninmathbb{Z}} int_{0}^a big|ncdot a+ybig>big<ncdot a+ybig| dy stackrel{?}{=}hat{1}_{mathcal{H}_d}otimes hat{1}_{mathcal{H}_c}
tag{4}
$$
where
$$
hat{1}_{mathcal{H}_c} =int_{0}^a big|ybig>big<ybig| dy tag{5}
$$
and
$$
hat{1}_{mathcal{H}_d}=sum_{ninmathbb{Z}} big|ncdot abig>big<ncdot abig| tag{6}
$$

From an intuitive point of view, I would expect the answer "yes", but I am looking for a rigorous proof (or disproof) of this. Reference to textbooks would be very appreciated.