Can 2-philosophers problem be analogous to dining philosophers problem?
Short answer: yes, though it could be considered a "degenerate case" (that is, a case that ends up being somewhat different from most others).

Long answer: the Dining Philosophers problem is really about having $n$ philosophers, and coming up with an algorithm that works for any $n$. A good algorithm should work for even "degenerate cases" like zero philosophers or one philosopher—by hard-coding them if necessary ("if there's only one philosopher, eat whenever you feel like"). So $n=2$ is a perfectly valid Dining Philosophers problem.