No: to make it true, you need to replace 1) with "$M_R$ is a Noetherian module."
The most obvious counterexample to the original statement you gave is to take any ring with identity that isn't right Noetherian. While the ring as a right module is cyclic, it must have a right ideal that isn't finitely generated.
On another note, if a ring has this property that finite generation is "closed downward" (or "inherited by submodules") for all right modules, then it can be shown the ring is right Noetherian. In fact, this characterizes right Noetherian rings.
No: to make it true, you need to replace 1) with "$M_R$ is a Noetherian module."
The most obvious counterexample to the original statement you gave is to take any ring with identity that isn't right Noetherian. While the ring as a right module is cyclic, it must have a right ideal that isn't finitely generated.
On another note, if a ring has this property that finite generation is "closed downward" (or "inherited by submodules") for all right modules, then it can be shown the ring is right Noetherian. In fact, this characterizes right Noetherian rings.