Here is a constructive proof (though I am not saying that the proofs given so far have been non-constructive; I have not tracked down their dependencies).

I will rely on the book Henri Lombardi, Claude Quitté, *Commutative algebra: Constructive methods. Finite projective modules*, arXiv:1605.04832v2 (translated by Tania K. Roblot, published by Springer 2015, but the link is to an updated version gracefully posted on the arXiv). Specifically, I will need the following fact from that book:

**(1)** If $R$ is a Prüfer ring (not necessarily a Prüfer domain!), and if $\mathfrak{a}$ and $\mathfrak{b}$ are two finitely generated ideals of $R$, then $\mathfrak{a} \mathfrak{b} = \left(\mathfrak{a} \cap \mathfrak{b}\right) \left(\mathfrak{a} + \mathfrak{b}\right)$.

This is part of Exercise 10 point 3 in Chapter VIII of the above book.

Here is your claim, slightly extended (as Laurent pointed out, it is not necessary to all-quantify over $b$ in the assumption):

**Proposition 1.** Let $R$ be a Prüfer domain. Let $a \in R$ and $b \in R$ be such that the ideal $Ra \cap Rb$ of $R$ is principal. Then, the ideal $Ra + Rb$ of $R$ also is principal.

*Proof of Proposition 1.* We WLOG assume that $a \neq 0$, since otherwise the claim is obvious (because $a = 0$ entails $Ra + Rb = Rb$). For similar reasons, we WLOG assume that $b \neq 0$. The ideal $Ra \cap Rb$ of $R$ is principal; in other words, $Ra \cap Rb = Rc$ for some $c \in R$. Consider this $c$. The element $ab$ of $R$ belongs to both $Ra$ and $Rb$, and thus belongs to $Ra \cap Rb = Rc$. In other words, $ab = cx$ for some $x \in R$. Consider this $x$. We shall show that $Ra + Rb = Rx$.

Indeed, **(1)** (applied to $\mathfrak{a} = Ra$ and $\mathfrak{b} = Rb$) yields $\left(Ra\right)\left(Rb\right) = \underbrace{\left(Ra \cap Rb\right)}_{=Rc} \left(Ra + Rb\right) = Rc \left(Ra + Rb\right) = c\left(Ra + Rb\right)$. Thus, $c\left(Ra + Rb\right) = \left(Ra\right)\left(Rb\right) = R\underbrace{ab}_{=cx} = Rcx = c\left(Rx\right)$.

But $R$ is an integral domain. Hence, from $a \neq 0$ and $b \neq 0$, we obtain $ab \neq 0$, so that $cx = ab \neq 0$ and thus $c \neq 0$. Therefore, if two ideals $I$ and $J$ of $R$ satisfy $cI = cJ$, then $I = J$ (to prove this, just argue elementwise, again using the fact that $R$ is an integral domain). Applying this to $I = Ra + Rb$ and $J = Rx$, we conclude that $Ra + Rb = Rx$ (since $c\left(Ra + Rb\right) = c\left(Rx\right)$). Hence, the ideal $Ra + Rb$ is principal. $\square$

*Remark.* The above proof still works if we replace the condition that $R$ be a domain by the condition that $a$ and $b$ be non-zero-divisors.