This is false.

**Example.** Let $A_i = A'_i = \mathbf Z^{(\mathbf N)}$ be an infinitely generated free abelian group. For $i < j$, let $f_{ij} \colon A_i \to A_j$ be the map that kills the first $j$ coordinates, and $f'_{ij} \colon A'_i \to A'_j$ the map that kills the even coordinates up to $2j$:
\begin{align}
f_i((a_j)_j) &= (0,\ldots,0,a_{j+1},\ldots),\\
f'_i((a_j)_j) &= (a_1,0,a_3,0,\ldots,a_{2j-1},0,a_{2j+1},a_{2j+2},\ldots).
\end{align}
It is clear that the $A_i$ and $A'_i$ are pairwise isomorphic (even in finite subsets). However, the colimits are not isomorphic:
$$A = \underset{\longrightarrow}{\operatorname{colim}} A_i = 0,$$
since an element $(a_1,\ldots,a_j,0,0,\ldots)$ in some $A_i$ disappears in $A_j$. On the other hand,
$$A' = \underset{\longrightarrow}{\operatorname{colim}} A'_i = \mathbf Z^{(2\mathbf N-1)}$$
consists of all finite sequences $(a_1,\ldots) \in \mathbf Z^{(\mathbf N)}$ supported on odd coordinates only. Indeed, the even coordinates all die by the same argument, and the odd ones never change in the colimit.