Given $H$ a presentable category and $S$ a set of maps in $H$ then the fullcategory $H^S$ of objects in $H$ that are right orthogonal to every arrow in $S$ is a reflective subcategory of $H$.

Moreover the reflexion $H \rightarrow H^S$ is the "cocontinuous localization of $H$ at $S$", meaning that it is universal among cocontinuous functor $H \rightarrow D$ which send every arrow in $S$ to an invertible arrow.

Indeed given a cocontinuous functor $H \rightarrow D$ it automatically has a right adjoint by presentability of $H$. Assuming that morphisms of $S$ are inverted in $D$, it is easy to see that this right adjoint takes values in $H^S$, which in turn implies that every morphisms in $H$ that is inverted by the reflection $H \rightarrow H^S$ is inverted by $H \rightarrow D$. It follows that the unit of the reflection $H \rightarrow H^S$ is inverted and hence that $H \rightarrow D$ is isomorphic to the composite $H \rightarrow H^S \rightarrow D$.

This imediately answer your question positively once you fix what I think are small problems with limit/colimits and completeness/cocompletness confusions.

The data of a functor $C \rightarrow D$ with $D$ a cocomplete category is the same as a cocontinuous functor $PC \rightarrow D$ by Kan extension. The condition that:

$$ colim F(C_u) \rightarrow F(X) $$

is invertible means that the maps from $C_u \rightarrow X$ in $P(C)$ is sent by (the extention of) $F$ to an invertible arrow, and hence by the discusion above, the functor $F$ factors through the reflexion on category of objects that are right orthogonal to all these maps... and this is exactly the definition of the category of sheaves.

This version has little to do with topos theory. The results you mentioned (for example in Lurie, though the 1-categorical version was known long before this) Is an additional step on these observation checking that if $C \rightarrow D$ preserve finite limits, that $D$ is a Grothendieck topos and that the localization $P(C) \rightarrow P(C)^S$ is left exact then the functor $Sh(C)=P(C)^S \rightarrow D$ that you get this way is also left exact.