The Problem
Let $\mathcal{C}$ be an $\infty$-category and $K$ a simplicial set. I believe that the following statement is true:
The full subcategory $\operatorname{Fun}^c(K^\triangleright,\mathcal{C})\subset \operatorname{Fun}(K^\triangleright,\mathcal{C})$ spanned by the colimiting cones is closed under colimits in $\operatorname{Fun}(K^\triangleright,\mathcal{C}).$
In other words, if $X$ is another simplicial set and $f:X\to\operatorname{Fun}(K^\triangleright,\mathcal{C})$ is a diagram which factors through $\operatorname{Fun}^c(K^\triangleright,\mathcal{C})$, and if $f$ has a colimit cone $\overline{f}:X^\triangleright\to\operatorname{Fun}(K^\triangleright,\mathcal{C})$, then the image $\overline{f}(\infty)$ of the cone point $\infty\in X^\triangleright$ should again be a colimiting cone. But I haven't been able to give a complete proof of this fact, and I need help. Below I indicate what I have tried. I appreciate any comments/suggestions/ideas. Thanks in advance.
Things I tried
I could prove the claim in the case $\mathcal{C}$ has colimits of shape $K$. It suffices to assume that $\mathcal{K}=K$ is an $\infty$-category by factoring $K\to \ast$ as an inner anodyne followed by an inner fibration. The restriction functor $\operatorname{Fun} ^c(\mathcal{K}^\triangleright,\mathcal{C})\to\operatorname{Fun}(\mathcal{K},\mathcal{C})$ is then a trivial fibration (by Proposition 4.3.2.15 of Lurie's HTT), and any section of it composes with the inclusion $\operatorname{Fun}^c(\mathcal{K}^\triangleright,\mathcal{C})\hookrightarrow \operatorname{Fun}(\mathcal{K}^\triangleright,\mathcal{C})$ to yield a left adjoint $i_!$ of the restriction functor $i^*:\operatorname{Fun}(\mathcal{K}^\triangleright,\mathcal{C})\to \operatorname{Fun}(\mathcal{K},\mathcal{C})$ (Proposition 4.3.2.17 of HTT). Since colimits in functor categories can be computed pointwise (Kerodon, Tag 02X9), the cone $i^*\overline{f}$ is still a colimit cone. Thus $i_! i^* \overline{f}$ is also a colimit cone. As an endo-functor of $\operatorname{Fun}^c(\mathcal{K}^\triangleright,\mathcal{C})$, the functor $i_!i^*$ is naturally equivalent to the identity functor, so the diagrams $i_! i^* f$ and $f$ are naturally equivalent. The former admits a colimiting cone $i_!i^*\overline{f}$ taking values in $\operatorname{Fun}^c(\mathcal{K}^\triangleright,\mathcal{C})$, so this completes the proof in the special case under consideration.
Let $\mathcal{S}$ be the $\infty$-category of Kan complexes, and let $j:\mathcal{C}^{\mathrm{op}}\to\operatorname{Fun}(\mathcal{C},\mathcal{S})$ denote the Yoneda embedding. The functor $j$ preserves limits (HTT, Proposition 5.1.3.2), so $j^{\mathrm{op}}$ preserves colimits. Therefore, the composite $$X^\triangleright \xrightarrow{\overline{f}} \operatorname{Fun}(K^{\triangleright},\mathcal{C})\xrightarrow{j^\mathrm{op}_\ast} \operatorname{Fun}(K^{\triangleright},\operatorname{Fun}(\mathcal{C},\mathcal{S})^\mathrm{op})$$ is still a colimit cone. Since $\operatorname{Fun}(\mathcal{C},\mathcal{S})^\mathrm{op}$ is cocomplete, the argument made in the question shows that the functor $j\overline{f}(\infty):K^\triangleright \to \operatorname{Fun}(\mathcal{C},\mathcal{S})$ is a colimit cone. But now $j$ is fully faithful (HTT, Proposition 5.1.3.1), so this implies that $\overline{f}(\infty):K^\triangleright \to\mathcal{C}$ is a colimit cone.
(Of course I should have used the Yoneda embedding, since that's how the corresponding claim in ordinary category theory is proved!)