Put $c = 1$ in the given condition, we have
$$f(ab + a + b) = f(a)f(b) + f(a)f(1) + f(b)f(1); \quad \forall a, b \in \mathbb{Q}^{+}, \quad (1)$$
Put $b = 3$ into (1), we have
$$f(4a + 3) = f(a)f(3) + f(a)f(1) + f(3)f(1); \quad \forall a \in \mathbb{Q}^{+}.$$
Put $b = 1$ into (1), we have
$$f(2a + 1) = 2f(a)f(1) + f(1)^2; \quad \forall a \in \mathbb{Q}^{+}.$$
Thus
$$f(4a + 3) = 2f(2a + 1)f(1) + f(1)^2 = 4f(1)^2f(a) + 2f(1)^3 + f(1)^2; \quad \forall a \in \mathbb{Q}^{+}.$$
From these, we can conclude that
$$[f(3) + f(1)] f(a) + f(3)f(1) = 4f(1)^2f(a) + 2f(1)^3 + f(1)^2; \quad \forall a \in \mathbb{Q}^{+}.$$
If $f(3) + f(1) \neq 4f(1)^2$ then $f$ is constant. Thus $f(3) + f(1) = 4f(1)^2$, otherwise $f$ will be constant. So we must have
$$f(3) + f(1) = 4f(1)^2 \text{ and } f(3)f(1) = 2f(1)^3 + f(1)^2.$$
Thus $f(3), f(1)$ are solutions of the quadratic equation $t^2 - 2f(1)t + 2f(1)^3 + f(1)^2 = 0$, thus
$$f(1)^2 - 4f(1)^2 + 2f(1)^3 + f(1)^2 = 0 \Leftrightarrow f(1)^2(f(1) - 1) = 0.$$
$$f(ab + a + b) = f(a)f(b) + f(a) + f(b); \quad \forall a, b \in \mathbb{Q}^{+} \quad (2)$$
Continue to put $b = 1$ and $b = 3$, we get
$$f(4a + 3) = 4f(a) + 3 \text{ and } f(2a + 1) = 2f(a) + 1.$$
Put $a = b = c = \frac{1}{3}$ into the given condition, $f(\frac{1}{3}) = 3f(\frac{1}{3})^2$ so $f(\frac{1}{3}) = \frac{1}{3}$.
Put $a = 2$ and $b = \frac{1}{3}$ into (2), we have $f(3) = f(2)f(\frac{1}{3}) + f(\frac{1}{3}) + f(2)$, thus $f(2) = 2$.
Put $b = c = 2$ into the given condition, $f(4a + 4) = 4f(a) + 4$; $\forall a \in \mathbb{Q}^{+}$ thus
$$4f(a) + 4 = f(4a + 4) = f\left(4\left(a + \frac{1}{4}\right) + 3\right) = 4f\left(a + \frac{1}{4}\right) + 3.$$
From these, we can conclude that $f(a + \frac{1}{4}) = f(a) + \frac{1}{4}$, thus
$$f(4a + 4) = 4f(a) + 4; \quad \forall a \in \mathbb{Q}^{+}.$$
Hence, by induction, one can show that $f(x + n) = f(x) + n$ for all positive integer $n$ and positive real number $x$; thus $f(n) = n, \quad \forall n \in \mathbb{Z}^{+}$.
Finally, put $b \to n$ and $a \to \frac{m}{n+1}$ for some $m, n \in \mathbb{Z}^{+}$ into (2), we get
$$f(m + n) = f(n)f\left(\frac{m}{n+1}\right) + f\left(\frac{m}{n+1}\right) + f(n) \to f\left(\frac{m}{n+1}\right) = \frac{m}{n+1}.$$
Thus $f(x) = x$ for all $x \in \mathbb{Q}^{+}$. It is easy to check this function satisfies the condition. $\square$