Skolem's paradox and models in set theory Say you have a model $M$ (of $\mathsf{ZF}$) containing a set $S$ that you know is countable outside the model but the enumeration is missing from the model so that $S$ appears to be uncountable in $M$. 
Is it possible to add to the model a bijection between $S$ and a set $S'$ that is uncountable outside the model? And similarly, is it possible to add to the model a bijection between $S$ and an uncountable cardinal, say $\aleph_1^M$?
 A: I assume by "model" you mean "model of set theory". In question two, I'll also assume that $M$ is transitive and countable, so we do not have to worry about whether generics over $M$ exist, which is a technical issue that I believe would distract us from the problem at hand. 
The answer to the first question is no: A bijection between $S$ and an uncountable set $T$ would give us that $S$ is uncountable, against your assumption (if $S$ is countable, the enumeration does not go "away" by adding more sets to $M$). 
The answer to the second question, on the other hand, is yes. The method of forcing allows you to do that, resulting in a larger model $M[f]$ where $f$ is a bijection in $M$ between $S$ and, say, $\aleph_1^M$. Of course, we have that $\aleph_1^{M[f]}>\aleph_1^M$. 

That said, we can of course have a situation where $M\in N$ and $N$ is a model of set theory, and we have in $M$ a set $T$ that $N$ believes is uncountable, although $T$ is truly countable in the outside world, and we could then add to $M$ a bijection between $S$ and $T$. This would, of course, destroy the fact that $T$ is uncountable. This can be achieved by forcing over $N$. We then obtain models $M[f]\subset N[f]$ with $T$ countable in both.
More complicated scenarios are also possible where the extension is not achieved via forcing. For example, if enough large cardinals exist, then $\aleph_1^L$ is countable in $V$, and there is a real $0^\sharp$ such that in $L[0^\sharp]$ we have that $\aleph_1^L$ is countable. The real $0^\sharp$ cannot be added by forcing. 
Or we can have countable (but not transitive) $M\preceq N$ where $N$ is an end extension of $M$, and it is easy to arrange that the transitive collapse $\hat N$ of $N$ would be a superset of the transitive collapse $\hat M$ of $M$ and see as countable some sets that in $\hat M$ were uncountable.

More interesting "paradoxes" can be achieved: If you are willing to relax the requirement that $M$ is transitive. For example, H. Putnam noticed that as a straightforward corollary of Shoenfield's absoluteness, we have that any real (even $0^\sharp$) belongs to an $\omega$-model of $V=L$.
If there are enough large cardinals (in the true universe), in $L$ there will be transitive countable models $M$ of $V=L[0^\sharp]$, and more. By adding cardinals on top, we will eventually reach a countable $\alpha$ such that $L_\alpha$ is a model of set theory, all of $M$ belongs to $L_\alpha$, and all the members of $M$ are countable in $L_\alpha$.
A: It is possible, but not without adding ordinals.
The reason is that this bijection would mean that the replacement axiom fails, because the image of a set is the whole model. But it is possible, under some circumstances, when we do add ordinals. Consider the following scenario:
Assume that we have $\kappa_0<\kappa_1$ two inaccessible cardinals, then we can take a countable elementary submode of $V_{\kappa_1}$, $M$. So $M$ thinks that there exists an inaccessible cardinal, so $N=V_{\kappa_0}^M$ is a countable model of $\sf ZFC$.
Now over $M$ we force to add a bijection between $\kappa_0^M$ and $\aleph_1^M$, call that function $G$. Now we can't add $G$ to $N$ directly, but if we also add the rest of $M$ (which is an end-extension of $N$) and then add $G$ then we have added this bijection.
It's all quite convoluted. I know. Let me try and sum up:


*

*We have $M$ which is a countable model in which there exists an inaccessible $\kappa$. $N=V_\kappa^M$ is also a model of $\sf ZFC$, and $M$ is an end-extension of $N$.

*Add to $M$ a bijection from $\aleph_1^M$ and $\kappa$. Call this model $M[G]$.

*Now we can't add $G$ to $N$ directly, but we can first go to $M$ and then add $G$.

