1. Let $P(x, y)$ be the assertion of the given functional equation:
$$f(x - f(y)) = f(x) + a \lfloor y \rfloor$$
for all $x, y \in \mathbb{R}$.

2. First, we consider the case $a = 0$. If $a = 0$, the equation simplifies to:
$$f(x - f(y)) = f(x)$$
This implies that $f$ is a constant function. For example, $f(x) = c$ for some constant $c$. Therefore, $a = 0$ is a valid solution.

3. Now, assume $a \neq 0$. We need to find all possible values of $a$ such that the functional equation holds.

4. Suppose $f(u) = f(v)$ for some $u, v \in \mathbb{R}$. Comparing $P(x, u)$ and $P(x, v)$, we get:
$$f(x - f(u)) = f(x) + a \lfloor u \rfloor$$
$$f(x - f(v)) = f(x) + a \lfloor v \rfloor$$
Since $f(u) = f(v)$, we have:
$$f(x - f(u)) = f(x - f(v))$$
Therefore:
$$f(x) + a \lfloor u \rfloor = f(x) + a \lfloor v \rfloor$$
This implies:
$$a \lfloor u \rfloor = a \lfloor v \rfloor$$
Since $a \neq 0$, we conclude:
$$\lfloor u \rfloor = \lfloor v \rfloor$$
Hence, if $f(u) = f(v)$, then $\lfloor u \rfloor = \lfloor v \rfloor$.

5. Next, assume $f(u) \notin \mathbb{Z}$. Then there exist $x_1, x_2$ such that $\lfloor x_1 \rfloor = \lfloor x_2 \rfloor$ but $\lfloor x_1 - f(u) \rfloor \neq \lfloor x_2 - f(u) \rfloor$.

6. From $P(x_1, u)$ and $P(x_2, u)$, we have:
$$f(x_1 - f(u)) = f(x_1) + a \lfloor u \rfloor$$
$$f(x_2 - f(u)) = f(x_2) + a \lfloor u \rfloor$$
Since $\lfloor x_1 \rfloor = \lfloor x_2 \rfloor$, we get:
$$f(x_1 - f(u)) = f(x_2 - f(u))$$
This implies:
$$\lfloor x_1 - f(u) \rfloor = \lfloor x_2 - f(u) \rfloor$$
which is a contradiction. Therefore, $f(x) \in \mathbb{Z}$ for all $x$.

7. Using induction on $n$, we show that:
$$f(x - n f(y)) = f(x) - a n \lfloor y \rfloor \quad \forall n \in \mathbb{Z}$$
For $n = 1$, this is just $P(x, y)$. Assume it holds for $n$. Then:
$$f(x - (n+1) f(y)) = f((x - n f(y)) - f(y)) = f(x - n f(y)) - a \lfloor y \rfloor$$
By the induction hypothesis:
$$f(x - n f(y)) = f(x) - a n \lfloor y \rfloor$$
Therefore:
$$f(x - (n+1) f(y)) = f(x) - a n \lfloor y \rfloor - a \lfloor y \rfloor = f(x) - a (n+1) \lfloor y \rfloor$$
Thus, the induction is complete.

8. Now, consider:
$$f(x - f(z) f(y)) = f(x) - a f(z) \lfloor y \rfloor \quad \forall x, y, z$$
Swapping $y$ and $z$ and subtracting, we get:
$$f(z) \lfloor y \rfloor = f(y) \lfloor z \rfloor \quad \forall y, z$$
Setting $z = 1$, we get:
$$f(x) = c \lfloor x \rfloor \quad \forall x$$
for some $c \in \mathbb{Z}$.

9. Plugging this back into the original equation, we get:
$$f(x - c \lfloor y \rfloor) = f(x) + a \lfloor y \rfloor$$
$$c \lfloor x - c \lfloor y \rfloor \rfloor = c \lfloor x \rfloor + a \lfloor y \rfloor$$
Simplifying, we get:
$$c \lfloor x \rfloor - c^2 \lfloor y \rfloor = c \lfloor x \rfloor + a \lfloor y \rfloor$$
Therefore:
$$-c^2 \lfloor y \rfloor = a \lfloor y \rfloor$$
Since $\lfloor y \rfloor \neq 0$ for all $y$, we get:
$$a = -c^2$$
Hence, the answer is:
$$a \in \{-n^2 \mid n \in \mathbb{Z}\}$$

The final answer is $\boxed{ a \in \{-n^2 \mid n \in \mathbb{Z}\} }$