[This is now an answer to the edited question(s), with some details added.My answer to the original question is kept at the very end.]
Firstly: The question is a good one, and it is not easy to find references onthis. I had spent too much time pondering about the failure of the double dualargument (see below) before I finally heard the arguement given in the lastsection below, indirectly from Fantechi, via Faber.
Assume $X$ is smooth projective.
Definition: An $S$-valued point in $M_I(X)$ is an $S$-flat coherent sheaf on$S\times X$, with stable fibres of rank one, and with determinant line bundleisomorphic to $\mathcal{O}_{S\times X}$, modulo isomorphism.
(I do not know if this is what Bridgeland meant, but to me this is resonablystandard.)
Comment: Stability for rank one means torsion free.
Existence: Let $M(X)$ be the (Simpson) moduli space for stable rank onesheaves. Then $M_I(X)$ is the fibre over $\mathcal{O}_X$ for the determinantmap $M(X) \to \mathrm{Pic}(X)$. This map sends a sheaf $I$ (stable rank onefibres) on $S\times X$ to the determinant line bundle $\det(I)$, and it istrivial as a point in $\mathrm{Pic}(X)$ if it is of the form $p^*L$ with$L\in\mathrm{Pic}(S)$. Then $I\otimes p^*L^{-1}$ is equivalent to $I$in $M(X)(S)$, and it has trivial determinant. This shows that $M_I(X)$indeed is a fibre of the determinant map.
Of course the determinant of an ideal $I_Y\subset \mathcal{O}_X$ is nontrivialif $Y$ is a non principal divisor, so you cannot map such ideals to $M_I(X)$.In any case, the ideal of a divisor, without the embedding, would onlyremember the linear equivalence class.
For brevity, let $\mathrm{Hilb}(X)$ be the part of the Hilbert schemeparametrizing subschemes $Y\subset X$ of codimension at least $2$.Then there is a natural map $F: \mathrm{Hilb}(X) \to M(X)$ that sends an ideal$I_Y\subset\mathcal{O}_{S\times X}$ to $I_Y$, forgetting the embedding. Since $Y$ isflat, so is $I_Y$, and its fibres are torsion free (by flatness again) of rankone. By the codimension assumption, the determinant of $I$ is trivial.
Theorem: $F$ is an isomorphism.
Comment: In the literature one sometimes finds the argument that if $I$ is arank one torsion free sheaf with trivial determinant, then $I$ embeds into itsdouble dual, which coincides with its determinant $\mathcal{O}_X$. Thisestablishes bijectivity on points. (For Hilbert schemes of points on surfacesthis is enough to conclude, since you can check independently that both$\mathrm{Hilb}(X)$ and $M_I(X)$ are smooth, and that the induced map on tangentspaces is an isomorphism.) I do not know how to make sense of this argument infamilies.
Sketch proof of theorem: The essential point is to show that every $I$ in$M_I(X)(S)$ has a canonical embedding into $\mathcal{O}_{S\times X}$ such that thequotient is $S$-flat.
Let $U\subset S\times X$ be the open subset where $I$ is locally free. Itscomplement has codimension at least $2$ in all fibres. By the trivialdeterminant assumption, the restriction of $I$ to $U$ is trivial. By codimension $2$,the trivialization extends to a map $I\to \mathcal{O}_{S\times X}$. This map is injective,in fact injective in all fibres: The restriction to eachfibre ${s}\times X$ is nonzero (as $U$ intersectsall fibres) and hence an embedding ($I$ is torsion free in fibres). Itfollows that the quotient is flat. There are some details to check, butthis is the main point, I think.
[End of new answer, here is the original one:]
(I have not studied Bridgeland's paper, so I do not know the intended meaning there.)
