There is a way to turn a Delzant polytope into a lattice polytope, but there is no natural or canonical way. Note that the Delzant condition implies that the normal vector to each facet (codimension 1 face) is integral. By letting the defining equation for the hyperplane containing each facet rational (by slightly translating hyperplanes), you can make the coordinates of all vertices rational without changing the combinatorial structure. Now the polytope is integral after scaling. As you see, this is far from being canonical.

The above operation changes the cohomology class represented by the symplectic form. So you can think of it as choosing a different symplectic form on the same smooth manifold. If you view a symplectic toric manifold as a symplectic reduction, you are taking a different regular value of the moment map.

Some people regard symplectic toric manifolds as smooth projective varieties, but strictly speaking, that is wrong. Any symplectic toric manifold is diffeomorphic to a smooth projective variety, but it may not be symplecally embedded into a projective space equipped with the Fubini-Study form $\omega_{FS}$. In fact, by Kodaira embedding theorem, a symplectic toric manifold $(M, \omega)$ can be (symplectically) embedded into $\left(\mathbb{P}^N, \omega_{FS}\right)$ for some $N$ if and only if $[\omega] \in H^2(M, \mathbb{Z})$. Even if we allow scalar multiples of $\omega_{FS}$, the class $[\omega]$ can only be a multiple of an integral one.

For an easy example, consider $(S^2 \times S^2, \sigma \oplus \lambda\sigma)$ where $\sigma$ is a volume form on $S^2$ and $\lambda>0$ is irrational. It can never be embedded into a projective space preserving the symplectic structure. You can take rational $\lambda$ to find an embedding, but then you are dealing with a different symplectic manifold.