Any finite modification of a computable function is computable, so the only instance of your phenomenon is the set $S=\mathbb{N}$, for which the semi-characteristic function is the same as the characteristic function.

Meanwhile, a modified version of your question is answered by the concept of a *simple* set, a c.e. set whose complement is infinite but contains no infinite c.e. set. Namely,

**Theorem.** The following are equivalent for
$S\subset\mathbb{N}$ having infinite complement:

$S$ is simple. That is, $S$ is c.e. and the complement
is infinite, but has no infinite c.e. subset.

The semi-characteristic function of $S$ is computable,
but every computable function $f_S$ between the semi-characteristic function
and the characteristic function differs only finitely from
the semi-characteristic function of $S$; that is, the domain of $f_S$
is $S$ plus finitely many elements.

Proof. If $S$ is simple, then any such computable $f_S$ can
have only finitely many extra values, since the zeros of
$f_S$ would be a c.e. set in the complement of $S$.
Conversely, if $S$ is c.e. but not simple, then you can
build a suitable $f_S$ by extending the semi-characteristic
function to include that extra infinite piece, on which
$f_S$ has value $0$. QED

Post had hoped to use such sets to solve Post's problem (the question of whether there are c.e. sets of strictly intermediate Turing degree complexity between the computable sets and the halting problem), and their investigation inspired a huge part of computability theory, even though the ultimate resolution of Post's problem didn't involve that concept.

characteristic functionis also completely standard for the function having value $1$ on $S$ and value $0$ on the complement of $S$, and this is the sense that is used here. See en.wikipedia.org/wiki/Characteristic_function. $\endgroup$ – Joel David Hamkins Dec 28 '10 at 21:25