Let $F$ be a sheaf which is trivial on $X - S$. This means that $j^*F = O_{X-S}^{\oplus n}$ for some $n$, where $j$ is the embedding of $X - S$ into $X$. Note that by Hartogs theorem one has $j_*O_{X-S} = O_X$, so by adjunction one has a morphism $F \to j_*j^*F = O_X^{\oplus n}$ which is an isomorphism on $X - S$. So, this means that such $F$ is a modification of the trivial bundle in the finite number of points. In other words, $Coh(X)_{codim 0}$ is the extenion closed abelian subcategory of $Coh(X)$ generated by $O_X$ and all structure sheaves of points.

Now indeed, as Will Sawin pointed out this category only remembers the dimension of $X$ and the cohomology of its structure sheaf --- given two varieties $X$ and $Y$ with the same dimension and cohomology of the structure sheaves one can construct an equivalence of these categories just by taking $O_X$ to $O_Y$ and choosing a bijection between the sets of points (and isomorphisms of tangent spaces for the corresponding points).

Let $F$ be a sheaf which is trivial on $X - S$. This means that $j^*F = O_{X-S}^{\oplus n}$ for some $n$, where $j$ is the embedding of $X - S$ into $X$. Note that by Hartogs theorem one has $j_*O_{X-S} = O_X$, so by adjunction one has a morphism $F \to j_*j^*F = O_X^{\oplus n}$ which is an isomorphism on $X - S$. So, this means that such $F$ is a modification of the trivial bundle in the finite number of points. In other words, $Coh(X)_{codim 0}$ is the extenion closed abelian subcategory of $Coh(X)$ generated by $O_X$ and all structure sheaves of points.

EDIT. Let us see what the information we have in this category. First, we have the cohomology of $O_X$. Second, we have a bunch of structure sheaves of points, and this part of the category depends only on the dimension of the variety. Finally, we have multiplication maps $$ Ext^\bullet(O_X,O_S)\otimes Ext^\bullet(O_S,O_X) \to Ext^\bullet(O_X,O_X) $$ and $$ Ext^\bullet(O_X,O_S)\otimes Ext^\bullet(O_S,O_X) \to Ext^\bullet(O_S,O_S). $$ Since $S$ is $0$ dimensional the first factor lives in degree $0$ and the second in degree $n = \dim X$, so the only thing we have are the maps $$ Hom(O_X,O_S)\otimes Ext^n(O_S,O_X) \to Ext^n(O_X,O_X) $$ and $$ Hom(O_X,O_S)\otimes Ext^n(O_S,O_X) \to Ext^n(O_S,O_S). $$ Now let $S= x$ be a point. By Serre duality these maps correspond to the maps $$ Hom(O_X,O_x)\otimes Hom(O_X,\omega_X) \to Ext^n(O_X,\omega_X\otimes O_x) $$ and $$ Hom(O_X,O_x)\otimes Hom(O_x,O_x) \to Hom(O_X,O_x). $$ So, the second map contains no information, while the first gives the canonical morphism $$ X \to P(H^0(\omega_X)^*) = P(H^n(O_X)). $$

So, on one hand, if $H^n(O_X) = 0$ you cannot reconstruct anything. On the other hand, if the canonical class is very ample, by reconstrcting the anticanonical image you reconstruct all $X$. I am not quite sure what goes on in the intermediate cases.