Substituting $x = y = 0$ in the functional equation, we get $f^2(0) = 1$ or $f(0) = \pm 1$.

If $f(0) = 1$ then substituting $x = x, y = 0$, we get $f(0) = f(x) \cdot f(0) + f(x) - 1$ or $f(x) = 1$. So in this case $f(x) = 1$ is a constant function.

If $f(0) = -1$, then substituting $x = y = 2$, we get $f(4) = f^2(2) + f(4) - 1$ or $f(2) = \pm 1$.

Now suppose that $f(2) = -1$. Then, substituting $x = y = 1$, we get $f(1) = f^2(1) + f(2) - 1 \Leftrightarrow f^2(1) - f(1) - 2 = 0$. From this quadratic equation we find that $f(1) = -1$ or $f(1) = 2$.

If $f(1) = 2$ then substituting $x = x$, $y = 1$ we get $f(x) = f(x)f(1) + f(x+1) - 1 \Leftrightarrow f(x+1) = 1 - f(x)$. From here we get $f(x+2) = 1 - f(x+1) = f(x)$. Also, another way, substituting $x = x$, $y = 2$ then we get $f(2x) = f(x)f(2) + f(x+2) - 1 = -f(x) + 2f(x+1) + 1 - 1 = -f(x) + 2f(x+1)$. But from above, $f(x+1) = 1 - f(x)$, so $f(2x) = -f(x) + 2(1 - f(x)) = -f(x) + 2 - 2f(x) = 2 - 3f(x)$. But for every $x \in \mathbb{Q}$ we get that $f(x) = -1$, but we have $x = 1$ for $f(1) = 2$. Which is a contradiction.

Now if $f(1) = -1$ then substituting $x = x$, $y = 1$ we get
$$f(x) = f(x) \cdot f(1) + f(x+1) - 1 \Leftrightarrow f(x+1) = 2f(x) + 1. \quad (1)$$

On the other hand, we get
$$\begin{aligned} f(2x) &= f(x) \cdot f(2) + f(x+2) - 1 = -f(x) + 2f(x+1) + 1 - 1 = \\ &= -f(x) + 2(2f(x) + 1) = -f(x) + 4f(x) + 2 = 3f(x) + 2. \end{aligned} \quad (2)$$

By using (1) we get that:
$$\begin{aligned} f(m) &= f(m-1) + 1 = f(m-2) + 2 = \dots = f(2) + m - 2 = m - 1. \end{aligned}$$
for every $m \in \mathbb{N}$, since $f(2) = 1$ (see below).

Now, let's check $f(2)$. If $f(2) = -1$, as above, we get a contradiction. So $f(2) = 1$.

Now substituting $x = y = 1$, we get $f(1) = f^2(1) + f(2) - 1$. So $f(1) = 0$ or $f(1) = 1$. If $f(1) = 1$ then substituting $x = x$, $y = 1$ we get $f(x) = f(x) \cdot f(1) + f(x+1) - 1$ from this $f(x+1) = 1$. Substituting $x = -1$ then $f(0) = 1$ which is a contradiction with $f(0) = -1$.

Now we have $f(0) = -1$, $f(1) = 0$, $f(2) = 1$. For every $x \in \mathbb{Q}$:
$f(x) = f(x) \cdot f(1) + f(x+1) - 1 \Leftrightarrow$
$$\Leftrightarrow f(x+1) = f(x) + 1. \qquad (3)$$

Then for every $m \in \mathbb{N}$
$$f(m) = f(m-1) + 1 = f(m-2) + 2 = \dots = f(2) + m - 2 = m - 1.$$

Also,
$$0 = f(1) = f\left(m \cdot \frac{1}{m}\right) = f(m) \cdot f\left(\frac{1}{m}\right) + f\left(m + \frac{1}{m}\right) - 1$$
By using (3)
$$\begin{align*} f\left(m + \frac{1}{m}\right) &= f\left(m - 1 + \frac{1}{m}\right) + 1 = \dots = f\left(\frac{1}{m}\right) + m \Leftrightarrow \\ \Leftrightarrow 0 &= (m - 1)f\left(\frac{1}{m}\right) + f\left(\frac{1}{m}\right) + m - 1 \end{align*}$$
So we get $f\left(\frac{1}{m}\right) = \frac{1}{m} - 1$.

Now for every $m, n \in \mathbb{N}$,
$$\begin{align*} f\left(\frac{n}{m}\right) &= f(n) \cdot f\left(\frac{1}{m}\right) + f\left(n + \frac{1}{m}\right) - 1 = \\ &= (n - 1) \cdot \left(\frac{1}{m} - 1\right) + n + \frac{1}{m} - 1 - 1 = \frac{n}{m} - 1. \end{align*}$$

Therefore, the solutions are: for every $x \in \mathbb{Q}$, $f(x) = -1$ or $f(x) = x - 1$.