Just saw this question and thought I would mention the ``explicit construction". You seem to be happy with reflexive coequalisers, as they are sifted colimits can just be computed pointwise in Prod(L,Set). From coproducts and reflexive coequalisers in a category one can get arbitrary coequalisers:
To give coequalisers in a category A is to give a left adjoint to $\Delta:A \to Graph(A)$, to give reflexive coequalisers is to give a left adjoint to $\Delta:A \to RGraph(A)$. But if you have coproducts then then the forgetful functor $RGraph(A) \to Graph(A)$ has a left adjoint, where you freely add identities to a graph $d,c:A \to B$ (two arrows meant), giving a new graph $d_{1},c_{1}:A+B \to B$ with the evident description. Therefore composing this left adjoint $Graph(A) \to RGraph(A)$ by the reflexive coequaliser functor $RGraph(A) \to A$ gives arbitrary coequalisers. Therefore reflexive coequalisers and coproducts gives coequalisers and coproducts and thus all colimits. The problem then is to understand coproducts. This is probably most easily seen on the monads side which again you seem happy with.
Any algebra for a monad is a coequaliser of frees, via its canonical presentation:
$(T^{2}A,\mu_{TA}) \to (TA,\mu_{A}) \to (A,a)$ (two arrows meant from $(T^{2}A,\mu_{A})$ to $(TA,\mu_{A})$ . This explicit presentation of an algebra contains all we need to construct an explicit presentation for coproducts. Well we must have that to give $(A,a)+(B,b) \to (C,c)$ is to give a pair $(A,a) \to (C,c)$ and $(B,b) \to (C,c)$ which, by the above presentation is to give maps $(TA,\\mu_{A}) \to (C,c)$ and $(TB,\\mu_{B}) \to (C,c)$ each coequalising those morphisms of the respective presentation. But to give $(TA,\\mu_{A}) \to (C,c)$ and $(TB,\\mu_{B}) \to (C,c)$ is to give an arrow from their sum, $(T(A+B),\mu_{A+B}) \to (C,c)$, the free algebra functor preserving coproducts. Now those maps of the respective presentations induce a pair $(T^{2}(A+B),\mu_{T(A+B)}) \to (T(A+B),\mu_{A+B})$ and to say that the maps $(A,a) \to (C,c)$ and $(B,b) \to (C,c)$ coequalise the previous pairs amounts to saying that the induced map $(T(A+B),\mu_{A+B}) \to (C,c)$ coequalises this pair; thus the coproduct of $(A,a)$ and $(B,b)$ must be the coequaliser of the maps $(T^{2}(A+B),\mu_{T(A+B)}) \to (T(A+B),\mu_{A+B})$ constructed (good exercise to work this out). This explains why the coproduct of a pair of groups and so on is a quotient of the free group on the disjoint union of the underlying sets. This construction may be interpreting just the same in the Lawvere theory setting.