The idempotent completion is idempotent complete and its inclusion is fully faithful and universal

By [L1], each hom-set of $\mathrm{Kar}(\mathcal{C})$ is the subgroup $f\mathcal{C}(A,B)e$ of the ambient hom-group, so it is an abelian group and composition is inherited bilinearly. Also $\mathrm{Kar}(\mathcal{C})((A,1_A),(B,1_B))=\mathcal{C}(A,B)$, so the inclusion $I$ is fully faithful and additive.

Let $u:(A,e)\to (A,e)$ be an idempotent in $\mathrm{Kar}(\mathcal{C})$. Then $u=eu=ue$ and $u^2=u$. The object $(A,u)$ of [L1] together with the morphisms $u:(A,e)\to (A,u)$ and $u:(A,u)\to (A,e)$ satisfies the splitting equations $u\cdot u=u$ on both sides, so $u$ splits. Hence $\mathrm{Kar}(\mathcal{C})$ is idempotent complete by [L2].

Define $\tilde{F}(A,e):=S_e$ and for $u:(A,e)\to (B,f)$ set $\tilde{F}(u):=p_{f}F(u)i_e$. This is well defined because $F(f)F(u)F(e)=F(u)$, so $i_f{p}_{f}F(u)i_e=F(u)i_e$ and $p_{f}F(u)i_e$ lands between the chosen splitting objects. For composable $u:(A,e)\to (B,f)$ and $v:(B,f)\to (C,g)$ one has $\tilde{F}(v)\tilde{F}(u)=p_{g}F(v)i_f{p}_{f}F(u)i_e=p_{g}F(v)F(f)F(u)i_e=p_{g}F(vu)i_e=\tilde{F}(vu)$, and the identity case is similar. Additivity follows from [L2].

For each object $A$ of $\mathcal{C}$, the identity pair $F(A)\stackrel{1}{\to }F(A)\stackrel{1}{\to }F(A)$ is a splitting of $F(1_A)=1_{F(A)}$. Since the supplied family also splits $F(1_A)$, [L3] gives a unique isomorphism ${\eta }_A:\tilde{F}(I(A))=S_{1_A}\to F(A)$ commuting with the two splittings. For a morphism $u:A\to B$, step 2.1 gives $\tilde{F}(I(u))=p_{1_B}F(u)i_{1_A}$, and the defining commutativities of ${\eta }_A$ and ${\eta }_B$ imply ${\eta }_B\tilde{F}(I(u))=F(u){\eta }_A$. So $\eta :\tilde{F}\text{\hspace{0.17em}⁣}\circ I\Rightarrow F$ is a natural isomorphism.

If a second splitting family is chosen, then [L3] gives for each object $(A,e)$ a unique isomorphism ${\theta }_e:S_e\to S_e^{\mathrm{\prime }}$ commuting with the two splittings of $F(e)$. For a morphism $u:(A,e)\to (B,f)$, the composites ${\theta }_f\tilde{F}(u)$ and ${\tilde{F}}^{\mathrm{\prime }}(u){\theta }_e$ are both maps $S_e\to S_f^{\mathrm{\prime }}$ compatible with those two splittings, so the same uniqueness forces ${\theta }_f\tilde{F}(u)={\tilde{F}}^{\mathrm{\prime }}(u){\theta }_e$. Hence the two extensions are naturally isomorphic.

If $\mathcal{C}$ is small and the Axiom of Choice is assumed, then one may choose a splitting of each idempotent $F(e)$ whenever $\mathcal{D}$ is idempotent complete. Applying steps 2.1-3.2 to such a choice gives the asserted universal extension.