We are tasked with finding the set of all $a \in \mathbb{R}$ for which there is no infinite sequence $(x_n)_{n \geq 0} \subset \mathbb{R}$ satisfying $x_0 = a$, and for $n = 0, 1, \ldots$, the equation

$$x_{n+1} = \frac{x_n + \alpha}{\beta x_n + 1}$$

is given with the condition $\alpha \beta > 0$.

First, consider the fixed points of the recurrence relation. A fixed point $x$ satisfies:

$$x = \frac{x + \alpha}{\beta x + 1}$$

Multiplying through by $\beta x + 1$ gives:

$$x(\beta x + 1) = x + \alpha$$

Simplifying, we have:

$$\beta x^2 + x = x + \alpha$$

$$\beta x^2 = \alpha$$

$$x^2 = \frac{\alpha}{\beta}$$

$$x = \pm \sqrt{\frac{\alpha}{\beta}}$$

Since we are interested in the set of $a$ for which the sequence cannot be infinite, these correspond to values where the iterations potentially stabilize and do not proceed infinitely.

Next, evaluate under the condition $\alpha \beta > 0$. This implies both $\alpha$ and $\beta$ have the same sign, which ensures that $\sqrt{\frac{\alpha}{\beta}}$ is a real number.

If $a = \sqrt{\frac{\alpha}{\beta}}$, then the sequence:

- Starts at $a$,
- Immediately lands on a fixed point, and
- Remains at this point, leading to failure in forming an infinite non-repetitive sequence, as it cycles at a constant value.

Consequently, the set of all $a$ for which there is no infinite sequence satisfying the given condition is precisely the fixed point we identified:

$$\boxed{\left\{ \sqrt{\frac{\alpha}{\beta}} \right\}}$$

Thus, the reference answer provided, being the set $\{ \sqrt{\frac{\alpha}{\beta}} \}$, is indeed correct.