1. Understanding the Problem:
We are given a figure $S$ in the plane such that its border does not contain any lattice points. We need to prove that if we can cover the plane with copies of $S$ such that any two lattice points $x$ and $y$ with distance 1 always go on lattice points, then the area of $S$ is equal to the number of lattice points inside it.

2. Initial Setup:
Let $S$ have $x \in \mathbb{N}$ lattice points in its interior and area $y$. We aim to show that $y = x$.

3. Contradiction Assumption:
Assume for contradiction that $y \neq x$. Select some $\epsilon > 0$ such that $y \notin [x(1 - \epsilon), x(1+\epsilon)]$.

4. Large Square Consideration:
Consider an arbitrarily large axis-aligned square of side length $a$ in the plane, where $a \in \mathbb{N}$. This square has approximately $a^2$ lattice points inside it.

5. Lattice Points in the Square:
Since $a$ is large, the number of lattice points inside the square is at least $(1 - \epsilon^2) a^2$ and at most $(1 + \epsilon^2) a^2$.

6. **Copies of $S$ in the Square:**
Each of these lattice points is contained in a copy of $S$. For at least $(1 - \epsilon^2)a^2$ of these lattice points, the copy of $S$ containing it is entirely within the $a \times a$ square.

7. Area Calculation:
The total area of the copies of $S$ inside the square is then at least $(1 - \epsilon^2)a^2 \cdot \frac{y}{x}$ and at most $(1 + \epsilon^2)a^2 \cdot \frac{y}{x}$.

8. Contradiction:
For small enough $\epsilon$ and large enough $a$, the range $[(1 - \epsilon^2)a^2 \cdot \frac{y}{x}, (1 + \epsilon^2)a^2 \cdot \frac{y}{x}]$ does not overlap with the range $[(1 - \epsilon^2)a^2, a^2]$. This leads to a contradiction.

9. Conclusion:
Therefore, our assumption that $y \neq x$ must be false. Hence, $y = x$.

$\blacksquare$