All $\mathsf{Set}$-valued presheaves on a cocomplete category are representable

No, it is not true. For example take the presheaf $F$ on $\mathsf{Set}$ itself (which is bicomplete, locally small, everything we could want) given by $F(X) = \{0,1\}$ for all sets $X$ (and all maps are mapped to the identity). Then $F$ is not representable, because there is no set $Z$ such that $\operatorname{Map}(X,Z)$ has two elements for all sets $X$.

The flaw in your argument can be made evident here: $F$ itself is the coproduct (as a functor!) of the representable presheaf $\operatorname{Map}(-, *)$ with itself, where $* \in \mathsf{Set}$ is some singleton, i.e. $F \cong Y(*) \sqcup Y(*)$. But $F$ is not isomorphic to $Y(* \sqcup *)$. The fact is, $Y$ does not preserve colimits: it preserves limits. It's not true that $\operatorname{Hom}(-, \operatorname{colim}_i X_i) \cong \operatorname{colim}_i \operatorname{Hom}(-, X_i)$, but it's true that $\operatorname{Hom}(-, \lim_i X_i) \cong \lim_i \operatorname{Hom}(-, X_i)$.

No. The Yoneda embedding preserves almost no colimits: the colimits you get are "free" and so have nothing a priori to do with any colimits that already exist. Work out explicitly what a coproduct of representable presheaves looks like: it is almost never representable.

Your argument cannot work for the simple reason that the category of presheaves is never (essentially) small. Also, Freyd showed that any cocomplete small category is a preorder, so even if your argument worked it would only apply, essentially, to suplattices.