I have not filled in absolutely all the details, but hopefully this is enough to be convincing.
Let's let $M$ take positive integer values, and let's consider the parallelepiped: $$P'=\{x+Mtv\mid x\in S, t\in [0,1]\}.$$ Because $v$ is a lattice vector and $S$ contains no lattice points, the number of lattice points in $P'$ is $M$ times the number of lattice points in $P$.
The volume of $P'$ is $M^3 (u\cdot v)$. For each lattice point $x$ in $P'$, consider a unit cube centred at $x$. "Most" of the lattice points in $P'$ are going to have the property that their entire cube lies in $P'$. More precisely, the difference between the volume of $P'$ and the sum of the volumes of the cubes centred at lattice points inside $P'$ is going to be bounded by something proportional to the area of the sides of $P'$ (because only lattice points near one of the boundary walls won't simply have their whole cube contained in $P'$). This area is of course quadratic in $M$. Another way to say this is that that the union of the unit cubes centred at the lattice points of $P'$ is a reasonable approximation to $P'$, with the difference in the two volumes being bounded by something quadratic in $M$.
Thus, the volume of $P'$ and the sum of the volumes of the cubes around the lattice points in $P'$ differ by an amount bounded by $cM^2$ for some $c$ depending on $S$ and $v$. The sum of the volumes of the unit cubes is, of course, the number of lattice points in $P'$.
Thus, $|M^3 (u\cdot v) - M|P\cap \mathbb Z^3| < cM^2$$|M^3 (u\cdot v) - M|P\cap \mathbb Z^3|| < cM^2$. Dividing through by $M^3$, we get the desired result.