Must a function that 'preserves r.e.-ness' be computable itself? Does there exist a non-recursive function (say, from naturals to naturals) such that the inverse of every r.e. set is r.e.?

If yes, how to construct one?
If no, how to prove that?
Any References?
For any sequence $\mathcal{B} = \langle B_i : i \in \mathbb{N}\rangle$ of subsets of $\mathbb{N}$, there is a $\mathcal{B}$-cohesive set $A$: an infinite set such that, for any $B_i \in \mathcal{B}$, either there are only finitely many elements of $A$ in $B_i$, or all but finitely many elements of $A$ are in $B_i$. The set $A$ is called "cohesive" because, although it is infinite, none of the sets $B_i$ is able to split it into two infinite pieces.

Apply that fact with $\mathcal{B}$ containing all the r.e. sets to obtain a cohesive set $A$. Let $f$ be the function that enumerates $A$ in increasing order.   First, $A$ is not computable, or even r.e., because if $A$ was r.e. then we could use that enumeration to effectively enumerate "every other element" of $A$, which would give us an r.e. set that splits $A$ into two infinite pieces, which is impossible.  Moreover, because $A$ is not computable, $f$ is not computable - if we could compute $f$ then we could enumerate $A$ in increasing order, which would imply $A$ is computable. 

Now let $B$ be any r.e. set. Then, because $A$ is cohesive for r.e. sets, one of two things must happen:


All but finitely many elements of $A$ are in $B$. In this case $f^{-1}(B)$ is cofinite, and hence is computable.
Only finitely many elements of $A$ are in $B$. In this case, $f^{-1}(B)$ is a finite set, and is thus computable.


Either way $f^{-1}(B)$ is not only r.e., it is computable.
I guess that you have meant image instead of inverse...

In that case in my opinion the answer is that such a function exists. Since you can construct a function $\phi$ as follows: for numbers representing valid pairs (code of a Turing machine, input) return 1, if the machine halts on the input, 0 otherwise. For all other numbers, return 0. This function is clearly not recursive. However, for all r.e. sets, the image is either $\{0\}$, or $\{0,1\}$, which are both r.e.