It seems like there are two many strange functions. The case $\alpha=\beta=1$ is in fact worse than Cauchy FE.

I am very sorry, but I made a typo on the RHS when copying. Now it should be correct...

$f(x)$ is bijective because can take any value and $f(x_1)=f(x_2) \to f(\alpha f(x_1)+f(y))=f(\alpha f(x_2)+f(y)) \to x_1=x_2$ $x=0: f(\alpha f(0)+f(y))=f(y) \to f(y)=y+c$ where $c=-\alpha f(0)$ is some const $f( \alpha f(x)+f(y))= \alpha x+y+ (\alpha +2)c = \beta x+ y+c$ so if $ \alpha \neq \beta$ then there are no solutions If $ \alpha = \beta = -1$ then $f(x)=x+c$ is solution for any $c$ If $\alpha = \beta \neq -1$ then $f(x)=x$ is solution

made mistake

Let $P(x,y)$ the assertion of the given F.E. As $\beta \ne 0$ we trivially have that $f$ is bijective (also note that $\alpha \ne 0$), therefore we have: $$f(\alpha f(x)+y)=\beta x+y \; \forall x,y \in \mathbb R$$Apply symetry and injectivity on this to $P(x, \beta y)$ we get $\alpha f(x)-\beta x=c$ therefore $f(x)=\frac{\beta x+c}{\alpha}$ so we must have $$f(\beta x+y+c)=\beta x+y \implies \beta^2 x+\beta y + \beta c + c=\alpha \beta x+\alpha y$$Now notice that by $x=y=0$ we get $c=0$ or $\beta=-1$, if $c=0$ then $\beta=\alpha$ and thus $f(x)=x$ which works, if $\beta=-1$ then we also have $\beta=\alpha=-1$ in which case $f(x)=x-c$ for any $c$ real which works as well, thus we are done .

Let $P(x,y)$ denote the assertion $f(\alpha f(x)+f(y))= \beta x+f(y)$. It is clear that $f$ is bijective. $P(0,x) \implies f(\alpha f(0)+f(x))=f(x) \implies \alpha f(0)+f(x)=x \implies f(x)=x-\alpha f(0)$. $P(0,0) \implies f(0)=-\alpha f(0) \implies (\alpha+1)f(0)=0 \implies f(0)=0$ or $\alpha=-1$. If $\alpha \neq -1$, then $f(x)=x$. $P(x,y) \implies \alpha x + y = \beta x + y \implies \alpha = \beta$. If $\alpha = -1$, then $f(x)=x+c$ where $c=f(0)$. $P(x,y) \implies f(-x-c+y+c)=\beta x + y + c \implies -x=\beta x \implies \beta=-1$. Hence the only solutions are $f(x)=x+c$ for some constant $c$ if $\alpha=\beta=-1$, $f(x)=x$ if $\alpha=\beta \neq -1$, and no solution for $\alpha \neq \beta$.

A different kind of FE lol, first time I've ever seen one of these. The answers are, 1. When $\alpha = \beta \neq -1$, $f(x)=x$ for all $x \in \mathbb{R}$ 2. When $\alpha = \beta =-1$, $f(x)=x+c$ for some constant $c\in \mathbb{R}$ for all $x \in \mathbb{R}$ 3. When $\alpha \neq \beta$ there exists no solutions. It is not hard to see that these functions are solutions to the given equation. Now, we shall show that they are the only ones. We denote by $P(x,y)$ the assertion that $f(\alpha f(x)+f(y))=\beta x+f(y)$ for given $\alpha , \beta \in \mathbb{R}$. Clearly $\alpha \neq 0$ as well since if $\alpha =0$, \[\beta x + f(y)=f(\alpha f(x)+f(y))=f(f(y))\]which is a clear contradiction since the left hand side can take multiple values. Now, from $P\left(\frac{-f(x)}{\beta},x\right)$ it follows that \[f(\alpha f\left(\frac{-f(x)}{\beta}\right)+f(x))= 0 \]from which it is clear that there exists $x_0 \in \mathbb{R}$ such that $f(x_0)=0$. Then, $P(x_0,x_0)$ gives us that \[f(0)=f(\alpha f(x_0)+f(x_0))=\beta x_0 f(x_0)=\beta x_0\]Thus, $f(0)=\beta x_0$. We can also note an important property of $f$. Claim : The function $f$ is injective. Proof : Say there exists real numbers $a \neq b$ such that $f(a)=f(b)$. Then, $P(a,y)$ and $P(b,y)$ give \[\beta b + f(y)=f(\alpha f(b)+f(y))=f(\alpha f(a)+f(y))=\beta a + f(y)\]from which it is clear that $a=b$, which proves the claim. Now, $P(0,x)$ gives us that \[f(\alpha \beta x_0 + f(x))=f(\alpha f(0)+f(x))=f(x)\]which since $f$ is injective implies $f(x)=x-\alpha \beta x_0$. Further, when $x=0$ we have $f(0)=-\alpha \beta x_0$ which implies $(\alpha+1)\beta x_0=0$ since $f(0)=\beta x_0$. Now, this implies that either $x_0=0$ or $\alpha =-1$. We examine these two possibilities separately. Case 1 : $\alpha \neq -1$ ($x_0=0$). Then, from $f(x)=x-\alpha \beta x_0$ we immediately have that $f(x)=x$ for all $x\in \mathbb{R}$. Plugging this into the given equation results in \[\alpha x + y =\beta x +y\]which is enough to imply $\alpha = \beta$. Case 2 : $\alpha =-1$. The given equation rewrites to \[f(f(y)-f(x))=\beta x + f(y)\]for all real numbers $x,y$. We denote this assertion by $Q(x,y)$. Then, $Q(0,x)$ gives \[f(f(x)-f(0))=f(x)\]which since $f$ is injective implies $f(x)=x+f(0)$ for all $x \in \mathbb{R}$. Plugging this into the original equation we have \[y-x + f(0)=\beta x + y + f(0)\]which is enough to imply that $\beta =-1$. Thus, in this case we must have $\beta =-1$ and all functions of the form $f(x)=x+c$ for a fixed constant $c$. Thus, we have checked all cases and indeed we have no additional solutions to the ones we claimed. We are done.