Let $R$ be a commutative ring with $1$ and $e$ be an idempotent element of $R$ with the property that if $e=x+y$ (where $x, y\in R$), then there exists $r\in R$ such that either $e=rx$ or $e=ry$. Can we deduced that for such an idempotent, the ideal $\langle e\rangle$ is a minimal ideal of $R$?

$\begingroup$ By a minimal ideal of $R$, do you mean a minimal nonzero ideal of $R$? Note that if $R = (\mathbb{Z}/4\mathbb{Z})^2$, then $e = (1, 0)$ satisfies your requirements. $\endgroup$ – Luc Guyot Jul 2 at 20:16
If you take $e=1$, then the condition imposed on $e$ translates to: for every $x \in R,$ either $x$ or $1x$ is a unit. This will be satisfied in any local ring $R$, while $R=(e)$ will not be minimal unless $R$ is a field. This already shows that the conclusion does not hold in general.
(If $e=1$ seems too trivial, this example should survive replacing the local ring $R$ by any ring of the form $R\times S$ and $e=1$ by $(1, 0)$).
On the other hand, if the Jacobson radical $J(R)$ of $R$ is $0$ and a nonzero idempotent $e$ has the imposed property, then $(e)$ is indeed minimal:
One needs to check that for $0 \neq x$ divisible by $e$, i.e. $ex=x$, one has $e$ divisible by $x$. In other words, if $x=ex$ is an element not dividing $e$, then $x=0$. The strategy is to show that $x$ lies in this case in the Jacobson radical, hence is $0$.
The imposed condition, together with the fact that $x$ does not divide $e$, implies that $(ex)=(eex)=e(1x)$ divides $e$. In particular, $(1x)$ divides $e$, and $e$ divides $x$, so $(1x)$ divides $x$. It also divides $(1x)$, so altogether $1x$ divides $1=(1x)+x$. Thus, $(1x)$ is a unit.
To finish the argument, note that $x$ in the above argument can be replaced by any multiple $x'=rx$. So we obtain that for all $r,$ $1rx$ is a unit, i.e. $x \in J(R)=0$.