Let $S$ be the set of pairs $(x,y)$ satisfying at least one of your bullet list. One can show that $S$ is an equivalence relation. Reflexivty and symmetry are almost immediate, and for transitivity assuming $(x,y),(y,z)$ lie in $S$ one can use the bulleted list to construct a chain from $x$ to $z$ using the assumed chains from $x$ to $y$ and from $y$ to $z.$
Now it's clear that since $R$ is the intersection of all equivalence classes containing r, and since $S$ is one of them, that $R \subseteq S.$ On the other hand $R$ must contain each pair $(x,y)$ in $S$ since $R$ contains $r$ and hence also your $r'.$ So $S \subseteq R$ and we get $R=S$ which shows what you wish to show.
Added note: Actually just from $R \subseteq S$ one can conclude what you want. (No need to use the full equality $R=S.$)