Let $\mathcal{T}$ be a Lawvere theory and let $\mathcal{C}$ be a category admitting finite products. Denote by $\mathrm{Mod}(\mathcal{T}, \mathcal{C})$ the category of models of $\mathcal{T}$ in $\mathcal{C}$.
Then, one can show that $\mathrm{Mod}(\mathcal{T}, \mathrm{Ind}(\mathcal{C}))$ admits small filtered colimits and the canonical embedding $\iota \colon \mathcal{C} \to \mathrm{Ind}(\mathcal{C})$ preserves finite products.
Since $\mathrm{Ind}$ is the free cocompletion with respect to filtered colimits, there is a unique (up to isomorphism) functor $\tilde{\iota_\ast} \colon \mathrm{Ind}(\mathrm{Mod}(\mathcal{T}, \mathcal{C})) \to \mathrm{Mod}(\mathcal{T}, \mathrm{Ind}(\mathcal{C}))$ such that $\tilde{\iota_\ast}$ preserves small filtered colimits and the diagram
commutes. In particular, for any $\alpha \colon I \to \mathrm{Mod}(\mathcal{T}, \mathcal{C})$ where $I$ is small and filtered, we have $\tilde{\iota_\ast}(\mathrm{''colim'' \alpha)} \cong \mathrm{colim} (\iota_\ast \circ \alpha)$.
Now, I was able to show that $\tilde{\iota_\ast}$ is fully faithful.
But e.g. for $\mathcal{C} = \mathbf{FinSet}$ and $\mathcal{T}$ the theory of groups, $\tilde{\iota_\ast}$ fails to be essentially surjective. In that case, the functor $\tilde{\iota_\ast}$ is given by
$\begin{matrix} \tilde{\iota_\ast} & \colon & \mathrm{Ind}(\mathbf{FinGrp}) & \to & \mathbf{Grp}\\ & & \mathrm{''colim''}(\alpha) & \mapsto & \mathrm{colim}(\alpha) \end{matrix}$
This functor is not essentially surjective since $\mathbb{Z}$ cannot be the colimit of finite groups.
Does anyone know a counterexample in the case $\mathcal{C} = \mathbf{Set}$?