Knowing a valid salted hash for an unknown secret, is it possible to compute another valid hash? $H(s,r)=d$ is a function that hashes the secret string $s$ with a salt $r$, and returns a digest $d$.
$r$ may be arbitrarily chosen and each $r$ returns a different $d$.
For any $d$, $r$ is known from a function $R(d)=r$.  

Both $H$ and $R$ are not reversible, i.e. there isn't a neither a reverse function $H'(d,r)=s$ nor a reverse $R'(r)=d$.
To prove that d is a valid hash for the string $s$, it's given a function $V(d,s) ↦ \{0,1\}$ that outputs $1$ only if $H(s,r)=d$.

Is it possible to compute another valid hash $d'$ for an unknown $s$, knowing only $d$ (and therefore the salt $r$)?
That is, is it possible to find a function $G(d,r')=d'$ such that $V(d',s)=1$ for at least some $r'≠ r$ matching $d'$?

This from a theoretical point of view.

On the practical side, BCrypt is an implementation of $H$. If a know a valid digest for an unknown secret, can I compute another valid digest? I've heard about a technique called "hash extension attack", but this involves changing the length of the digest, that for BCrypt is fixed. Is BCrypt vulnerable from this point of view? For which (if not all or none) cryptohashing algorithms is it possible to have a function like G?
Let's get terminology right. If you talk of "unknown s" then s is not a salt; when some piece of data is secret, we call it a key. And your "hash function" is then a MAC. In the context of "password hashing", such things are sometimes called "peppering" (as always, technical terminology is, at its core, a collection of bad puns).

If your MAC is correct (i.e. "cryptographically secure"), then no, it is not possible to produce new pairs message+hash value without knowing the key; being able to do that would be called a forgery. The length extension attack is a forgery attack on some poorly-designed MAC algorithms, which work by simply hashing the concatenation of the key and the message, with a Merkle-Damgård hash function like SHA-256. Don't take it wrong: the "length extension attack" does not mean that SHA-256 is cryptographically weak as a hash function; it just means that making a MAC out of a hash function is not as immediate as it initially seems to be. The proper way to turn a hash function into a MAC is HMAC.

Bcrypt is not a MAC, and, arguably, not a hash function either, since it does not accept as input a sequence of bits of arbitrary length. The salt in bcrypt is not secret at all; it is often encoded in the bcrypt output; moreover, if you are using the salt as a secret for some passwords, then you are reusing the same salt value for several hashed passwords, and that's very wrong. The whole point of the salt is to be distinct for every instance (every user has his salt; when the user changes his password, a new salt must be generated).

On the concept of password hashing, read this.



Edit: it has been brought to my attention that I misread the question somehow: the salt is known to be public. The question being: can we transform a hash for some secret password with some salt value, into another hash for the same secret password, but with another salt value ?

If such a thing is feasible and the new salt value can be arbitrarily chosen, then this would be a severe flaw in the hashing function, because it would allow the attacker to transform a bunch of hashed passwords into another bunch of hashes of the same passwords but all with the same salt -- thereby nullifying the advantage of the salt: parallel and precomputed attacks are again possible. Even if the attacker could just turn the salt into another "random" salt, then it could be used to make salt collisions more probable. No such flaw is currently known for bcrypt.