Replacing $(x, y)$ with $(y, x)$ shows we can assume $\alpha \leq \beta$, which gives $2\beta \geq \alpha + \beta \geq 2$ by the condition so $\beta \geq 1$. Taking $(0, 0)$ we obtain $f(0)^2 \leq f(0)$ so $0 \leq f(0) \leq 1$. First, assume that $f(0) = 0$. However, $(0, -1)$ then gives $0 \leq -\beta \leq -1$, a contradiction. Therefore $f(0) > 0$. Now, take $(0, y)$ to obtain $f(0)f(y) \leq f(0) + \beta y$ which rearranges to \[f(y) \leq \frac{\beta}{f(0)}y + 1\]for all real $y$. Now, take some $t \geq f(0)/\beta > 0$ and plug in $(-t, -t)$. We obtain: \[\frac{\beta^2}{f(0)^2}t^2 - \frac{2\beta}{f(0)}t + 1 = \left(\frac{\beta}{f(0)}t - 1\right)^2 \leq f(-t)^2 \leq f(t^2) - (\alpha + \beta)t \leq \frac{\beta}{f(0)}t^2 - (\alpha + \beta)t + 1\]for all $t$ sufficiently large, which implies $\beta^2/f(0)^2 \leq \beta/f(0)$ so we obtain $1 \leq \beta \leq f(0) \leq 1$ and we must have $1 = \beta = f(0)$ and $1 = \beta \geq \alpha \geq 1$ gives $\alpha = 1$. The chain of inequalities above now reduces to (for $t \geq 1$) \[t^2 - 2t + 1 \leq f(-t)^2 \leq f(t^2) - 2t \leq t^2 - 2t + 1\]so we obtain $f(t^2) = t^2 + 1$ and $f(-t)^2 = (t-1)^2$ for all $t \geq 1$. This gives $f(-1) = 0$ and taking $(-x, -1)$ we obtain $f(x) \geq x+1$ for all real $x$ which combined with the first inequality we proved gives $f(x) = x+1$ for all $x$, which is indeed a solution.

@axiomatical I would just add in a summary that there are no solutions for any $(\alpha,\beta)\ne(1,1)$

Here is a shorter solution: $P(-1,-1)$ shows $f(1) \ge f(-1)^2+\alpha+\beta \ge 2$. $P(1,1)$ shows $f(1)^2 \le f(1)+\alpha+\beta \le 2f(1)$. Hence $f(1) \le 2$ and hence $f(1)=2$ and hence $\alpha+\beta=2$ and $f(-1)=0$. $P(x,1)$ shows $f(x) \le \alpha x+\beta$. $P(x,-1)$ hows $f(x) \ge \alpha x+\beta$. Hence $f(x)=\alpha x+\beta$ but of course also $f(x)=\beta x+\alpha$ by symmetry. Hence $\alpha=\beta=1$ and $f(x)=x+1$ which is indeed a solution.