We have $f(a)$ and $f(b)$ have different signs. Without loss of generality $f(a)$ is positive and $f(b)$ is negative. Now choose $\epsilon>0$ sufficiently small such that $f(a)>\frac{\epsilon}{\epsilon+1}>-\epsilon>f(b)$. This is possible since as $\epsilon \to 0$, $\frac{\epsilon}{\epsilon+1} \to 0$ so the fraction can be made as small as possible and similarly, $-\epsilon$ can be made as large as possible less than $0$. Now by the Intermediate Value property, set $$f(\alpha)=\frac{\epsilon}{\epsilon+1}, \quad f(\beta)=-\epsilon$$Remark that $$f(\alpha)+f(\beta)=\frac{-\epsilon^2}{\epsilon+1}=f(\alpha)f(\beta)$$

Since $f(a)f(b)<0$ it follows that either $f(a)<0$ and $f(b)>0$, or $f(a)>0$ and $f(b)<0.$ We can assume WLOG that we are in the first case. Since $f$ has IVT, it follows that $[f(a),f(b)] \subset \mathrm{Im}(f).$ Thus $ \left( -\infty, \frac{1}{f(a)} \right) \cup \left( \frac{1}{f(b)}, \infty \right) \subset \frac{1}{f}(A),$ where $A= \{ x \in (a,b) : f(x) \neq 0 \}.$ We can find $u \in \left( -\infty, \frac{1}{f(a)} \right)$ such that $1-u \in \left( \frac{1}{f(b)}, \infty \right)$ so there exist $x,y \in A$ such that $\frac{1}{f(x)}=u$ and $\frac{1}{f(y)}=1-u.$ We can choose $\alpha = \min \{x,y\}$ and $\beta= \max\{x,y\}$ and we are done.

Romania 2021 11/1 wrote: Let $f:[a,b] \rightarrow \mathbb{R}$ a function with Intermediate Value property such that $f(a)f(b) < 0$. Show that there exist $\alpha$, $\beta$ such that $a < \alpha < \beta < b$ and $f(\alpha) + f(\beta) = f(\alpha) f(\beta)$. Since $f$ has the Intermediate Value Property, there exists $c$ such that $f(c) = x$ for any $x \in (f(a),f(b))$. Therefore, we can just rephrase this as Let $x,y$ be two real numbers such that $xy < 0$, (WLOG $y > x$) prove that there exists $i,j \in (x,y)$ such that $i + j = ij$. Note that $\left( \alpha, \frac{\alpha}{\alpha - 1} \right)$ is a solution to $i + j = ij$ for any $i,j$. Now, since $x < 0$ and $y > 0$, we could take $\varepsilon$ to be arbitrarily close to $0$ such that \[ i = \frac{\varepsilon}{\varepsilon + 1} , j = - \varepsilon \]which both tends to $0$ by taking $\varepsilon$ sufficiently small. Therefore, for a suitable $\varepsilon$, we could always choose it such that $i,j \in (x,y)$ since $x < 0$ and $y > 0$.