It's funny that you defend your view using Haskell, as I was replying with Haskell in mind.
If you use the state monad, you encode local side effects. Then you use runState in a function that has a non-monadic type--a pure function.
Feel free to argue that the state monad is only having pure expressions because it can be represented using only pure functions and partial evaluation. For all practical purposes it is used to write code that exerts (local) side effects into the state maintained by the state monad.
You can a function in say, OCaml, that fills an array via mutation using only function arguments to deduct the contents, and then returns that array. That function is pure (it exerts no side effects outside and is not exposed to side effects from the outside, its result value only depends on the function argument values). Yet how is that not using mutation, non-referentially transparent parts inside?
You can write the same function with a mutable vector in Haskell (Data.Vector.Mutable, using the ST monad). How is it not exactly the same code as the one in OCaml except for the types that guarantee the purity? Yes, it may be using pure expressions underneath to guide the type system to enable the proof of purity. But does that change what I said? Both to the programmer and (after optimization) to the machine the code is using mutation. You can model mutation using only pure functions. That may be useful in some ways, but you can also take a C program, represent every memory address as an entry in a tree of immutable nodes, the outside world as an abstract entity with versions, and recreate it with only pure expressions. That may be what is useful in some contexts for developing proofs or to implement type systems, but is this relevant for a programmer interested in writing more functional programs? Why not just say that using ST is using local mutation (that the type system can guarantee will not leak outside) and be done with it?