Functional equation $f(x+y)-f(x)-f(y)=alpha(f(xy)-f(x)f(y))$ is solvable without regularity conditions

I was reviewing this question and got motivated to solve this general problem:

Find all functions $f:\mathbb R\to\mathbb R$ such that for all real numbers $x$ and $y$,
$$f(x+y)-f(x)-f(y)=\alpha(f(xy)-f(x)f(y))\tag0$$
where $\alpha$ is a nonzero real constant.

I found out that it can be solved in a similar way to what I did in my own answer to that question. It was interesting for me that we don't need any regularity conditions like continuity. As I didn't find any questions about the same functional equation on the site, I thought it might be useful to post my own answer to it.

I appreciate any other ways to solve the problem or any insights helping me understand why there's no need of regularities while some functional equations like Cauchy's need such additional assumptions to have regular solutions.

Answer

First, letting $y=1$ in (0) and rearranging the terms, we get:
$$f(x+1)=f(1)+(1+\alpha-\alpha f(1))f(x)\tag1$$
$$\therefore\:f(2)=(2+\alpha-\alpha f(1))f(1)\tag2$$
Next, substituting $x+1$ for $x$ in (1) and using (1) and (2) we have:
$$f(x+2)=f(2)+(1+\alpha-\alpha f(1))^2f(x)\tag3$$
Next, letting $y=2$ in (0) and using (3) we get:
$$\left((1+\alpha-\alpha f(1))^2-1\right)f(x)=\alpha(f(2x)-f(2)f(x))$$

$$\therefore\:f(2x)=\left(2(1-f(1))+\alpha(1-f(1))^2+f(2)\right)f(x)$$
Thus using (2) we have:
$$f(2x)=(2+\alpha-\alpha f(1))f(x)\tag4$$
Now, substituting $2x$ for $x$ and $2y$ for $y$ in (0) and using (4), we'll get:
$$\beta(f(x+y)-f(x)-f(y))=\alpha\beta^2(f(xy)-f(x)f(y))$$
where $\beta=2+\alpha-\alpha f(1)$. Mutiplying (0) by $\beta$ and subtracting the last equation, we'll have:
$$\beta(\beta-1)(f(xy)-f(x)f(y))=0\tag5$$
If $\beta=0$ then by (4) we conclude that $f$ is the constant zero function. So this case can only happen when $\alpha=-2$.

If $\beta=1$ then by (1) we conclude that $f$ is the constant $1+\frac1\alpha$ function.

If $\beta\neq0$ and $\beta\neq1$ then by (0) and (5) we have:
$$f(xy)=f(x)f(y)\tag6$$
$$f(x+y)=f(x)+f(y)\tag7$$
By letting $y=x$ in (6) we conclude that $f(x)$ is nonnegative for nonnegative $x$. Combining with (7) we find out that $f$ is increasing. An increasing additive function is of the form $f(x)=kx$. So by (6) we conclude that $f$ is the constant zero function or the identity function.