If $\mathcal C$ has amalgamation, does $Ind(\mathcal C)$ have amalgamation? Recall that a category $\mathcal C$ has amalgamation if every span admits a cocone. If $\mathcal C$ has amalgamation, then does $Ind(\mathcal C)$ have amalgamation?
The "obvious way to show this" would be to induct on the presentability ranks of objects being amalgamated, presenting them as colimits of chains of objects of lower presentability rank. But this doesn't work -- it seems one can only prove amalgamation over finitely-presentable objects this way (i.e. for a span $B \leftarrow A \to C$ where $A$ is finitely-presentable, there is a cocone). EDIT and maybe not even that! See the comments
So I suspect the answer is no. But I don't have a counterexample.
Note that if $\mathcal C$ has functorial amalgamation, then something like the above proof attempt should work to show that $Ind(\mathcal C)$ does too. But I don't want to assume that amalgamation is functorial.
 A: A counterexample is given in the paper Disjoint Amalgamation in Locally Finite AEC by Baldwin, Koerwien, and Laskowski  (link). They're actually interested in finding AECs in which amalgamation holds for models up to size $\aleph_k$ but not for models of size $\aleph_{k+1}$ for $k\in \omega$. Since you're interested in (essentially) a special case of this,  I'll give a slightly simpler version of their construction, which is itself based on an earlier construction by Laskowski and Shelah (link). Of course, there might be a much simpler purely categorical example.
Consider the first-order language $L = \{f_n\mid n\in \mathbb{N}\}$, where the $f_n$ are binary function symbols. Let $K$ be the class of all finite $L$-structures which contain no independent subset of size $3$, i.e. such that there is no triple $(a,b,c)$ such that $a\notin \langle b,c\rangle$, $b\notin \langle a,c\rangle$, and $c\notin \langle a,b\rangle$, where $\langle X\rangle$ is the substructure generated by $X$.
Now consider the category $\mathcal{C}$ whose objects are $K$ and whose arrows are embeddings between structures in $K$.
Claim: $\mathcal{C}$ has the amalgamation property.
Suppose $f\colon A\to B$ and $g\colon A\to C$ are embeddings between structures in $K$, and identify $A$ with $f(A)\subseteq B$ and $g(A)\subseteq C$. Let $D$ be the disjoint union of $B$ and $C$ over $A$. To make $D$ into an $L$-structure we just need to define $f_n(b,c)$ when $b\in B\setminus A$ and $c\in C\setminus A$. Enumerate $D$ as $d_0,\dots,d_m$, set $f_n(b,c) = d_n$ for $n\leq m$ and $f_n(b,c) = b$ for $n>m$. Then $D\in K$, since for any triple $(d_1,d_2,d_3)$ either all $d_i$ are contained in $B$ or all $d_i$ are contained in $C$, in which case the triple is not independent by hypothesis, or we have WLOG $d_1\in B\setminus A$ and $d_2\in C\setminus A$, in which case $d_3\in \langle d_1,d_2\rangle$ by construction.
Claim: $\mathsf{Ind}(\mathcal{C})$ does not have the amalgamation property.
We can identify objects in $\mathsf{Ind}(\mathcal{C})$ with $L$-structures such that every finitely generated substructure is in $K$ and arrows in $\mathsf{Ind}(\mathcal{C})$ with embeddings between such structures. 
Now it's possible to find an infinite structure $A$ and extensions $B$ and $C$, all in $\mathsf{Ind}(\mathrm{C})$, with elements $b\in B\setminus A$ and $c\in C\setminus A$, such that $\langle Ab\rangle\subseteq B$ is not isomorphic over $A$ to a substructure of $C$, and similarly $\langle Ac\rangle\subseteq C$ is not isomorphic over $A$ to a substructure of $B$. That is, the quantifier-free type of $b$ over $A$ is not realized in $C$, and vice versa. 
Now suppose there is an amalgamation $D$ of $B$ and $C$ over $A$, and pretend all embeddings are inclusions. Then we must have $b\in D\setminus C$ and $c\in D\setminus B$. Since $\langle b,c\rangle$ is finite, there is some $a\in A\setminus \langle b,c\rangle$. Then $(a,b,c)$ is independent. Indeed, $\langle a,b\rangle\subseteq B$, so $c\notin \langle a,b\rangle$, and $\langle a,c\rangle\subseteq C$, so $b\notin \langle a,c\rangle$. It follows that $\langle a,b,c\rangle\notin K$, so $D$ is not in $\mathsf{Ind}(\mathcal{C})$, which is a contradiction.
