Let $\lfloor \bullet \rfloor$ denote the floor function. For nonnegative integers $a$ and $b$, their bitwise xor, denoted $a \oplus b$, is the unique nonnegative integer such that
$$\left \lfloor \frac{a}{2^k} \right \rfloor + \left \lfloor \frac{b}{2^k} \right \rfloor - \left \lfloor \frac{a \oplus b}{2^k} \right \rfloor$$
is even for every $k \ge 0$. We aim to find all positive integers $a$ such that for any integers $x > y \ge 0$, we have
$$x \oplus ax \neq y \oplus ay.$$

To solve this, we consider two cases based on the parity of $a$:

1. **Case 1: $a$ is even**

Suppose $a$ is even. Then $a = 2m$ for some integer $m$. Consider $x$ and $y$ such that $x > y \ge 0$. We have:
$$x \oplus ax = x \oplus 2mx$$
$$y \oplus ay = y \oplus 2my$$

Since $a$ is even, $2m$ is also even. The bitwise xor operation with an even number does not change the parity of the number of 1's in the binary representation. Therefore, $x \oplus 2mx$ and $y \oplus 2my$ will have the same parity, leading to a contradiction. Hence, for $a$ even, it is not possible to satisfy the given condition.

2. **Case 2: $a$ is odd**

Suppose $a$ is odd. Then $a = 2m + 1$ for some integer $m$. Consider $x = 2^k - 1$ and $y = 2^k + 1$ for sufficiently large $k$. We have:
$$x \oplus ax = (2^k - 1) \oplus (2m + 1)(2^k - 1)$$
$$y \oplus ay = (2^k + 1) \oplus (2m + 1)(2^k + 1)$$

Since $a$ is odd, the bitwise xor operation with an odd number will change the parity of the number of 1's in the binary representation. Therefore, $x \oplus ax$ and $y \oplus ay$ will not have the same parity, satisfying the given condition.

Thus, the only positive integers $a$ that satisfy the given condition are the odd integers.

The answer is: \boxed{\text{All odd positive integers}}.