Is 'Chaining operations' the "only" thing that the Monad class solves?
You're definitely on to something in the way that you're stating this—there are many things that Monad means and you've separated them out well.
That said, I would definitely say that chaining operations is not the primary thing solved by Monads. That can be solved using plain Functors (with some trouble) or easily with Applicatives. You need to use the full monad spec when "chaining with choice". In particular, the tension between Applicative and Monad comes from Applicative needing to know the entire structure of the side-effecting computation statically. Monad can change that structure at runtime and thus sacrifices some analyzability for power.
To make the point more clear, the only place you deal with a Monad but not any specific monad is if you're defining something with polymorphism constrained to be a Monad. This shows up repeatedly in the Control.Monad module, so we can examine some examples from there.
sequence :: [m a] -> m [a]
forever :: m a -> m b
foldM :: (a -> b -> m a) -> a -> [b] -> m a
Immediately, we can throw out sequence as being particular to Monad since there's a corresponding function in Data.Traversable, sequenceA which has a type slightly more general than Applicative f => [f a] -> f [a]. This ought to be a clear indicator that Monad isn't the only way to sequence things.
Similarly, we can define foreverA as follows
foreverA :: Applicative f => f a -> f b
foreverA f = flip const <$> f <*> foreverA f
So more ways to sequence non-Monad types. But we run into trouble with foldM
foldM :: (Monad m) => (a -> b -> m a) -> a -> [b] -> m a
foldM _ a [] = return a
foldM f a (x:xs) = f a x >>= \fax -> foldM f fax xs
If we try to translate this definition to Applicative style we might write
foldA :: (Applicative f) => (a -> b -> f a) -> a -> [b] -> f a
foldA _ a [] = pure a
foldA f a (x:xs) = foldA f <$> f a x <*> xs
But Haskell will rightfully complain that this doesn't typecheck--each recursive call to foldA tries to put another "layer" of f on the result. With Monad we could join those layers down, but Applicative is too weak.
So how does this translate to Applicatives restricting us from runtime choices? Well, that's exactly what we express with foldM, a monadic computation (a -> b -> m a) which depends upon its a argument, a result from a prior monadic computation. That kind of thing simply doesn't have any meaning in the more purely sequential world of Applicative.
To solve the problem of chaining operations on an individual monadic type, it's not at all necessary to make it an instance of Monad and be sure the monad laws are satisfied. You could just implement a chaining operation directly on your type.
It would probably be very similar to the monadic bind, but not necessarily exactly the same (recall that bind for lists is concatMap, a function that exists anyway, but with the arguments in a different order). And you wouldn't have to worry about the monad laws, because you would have a slightly different interface for each type, so they wouldn't have any common requirements.
To ask what problem the Monad type class itself solves, look at all the functions (in Control.Monad and else where) that work on values in any monadic type. The problem solved is code reuse! Monad is exactly the part of all the monadic types that is common to each and every one of them. That part is sufficient on its own to write useful computations. All of these functions could be implemented for any individual monadic type (often more directly), but they've already been implemented for all monadic types, even the ones that don't exist yet.
You don't write a Monad instance so that you can chain operations on your type (often you already have a way of chaining, in fact). You write a Monad instance for all the code that automatically comes along with the Monad instance. Monad isn't about solving any problem for any single type, it's about a way of viewing many disparate types as instances of a single unifying concept.