Suppose we have a preordered set $(I,\le)$ and a sequence of abelian groups $\{G_i\}$ with a homomorphism $\alpha_{i,j}:G_i\rightarrow G_j$ if $i\le j$ in $I$.
Let $G$ be the direct limit of this system.
If we remove some finite number of groups $G_i$, does it change the direct limit?
Same question I would concern for inverse limit of abelian groups.
It really depends on the order type of $I$.
For a trivial example, consider the set $I=\{1,2\}$ with its usual order, and the groups $G_1=\mathbb{Z}/2\mathbb{Z}$, $G_2=\mathbb{Z}/4\mathbb{Z}$, with $\alpha_{1,2}$ the unique embedding. The direct limit is just $G_2$; if you delete $1$, you still get $G_2$; but if you delete $2$, then you get $G_1$, which is not isomorphic to $G_2$. Likewise with inverse limits. (You can construct examples with any infinite set that has a maximum, since the direct limit over a set with a maximum is just the object indexed by the maximum.)
What you really want to consider is whether the remaining index set is cofinal in $I$.
Definition. Let $I$ be a pre-ordered ordered set, and let $J$ be a subset of $I$. We say that $J$ is cofinal in $I$ if and only if for every $i\in I$ there exists $j\in J$ such that $i\leq j$.
Theorem. Let $I$ be a preordered set. Let $\{G_i\}_{i\in I}$ be a family of (abelian) groups, and let $\{\alpha_{ij}\}_{i\leq j}$ be a family of compatible homomorphisms. If $J$ is cofinal in $I$, then $$\lim_{\longrightarrow I}G_i \cong \lim_{\longrightarrow J}G_j.$$
The proof is given here. A similar result holds for inverse limits.
If your set $I$ is directed and has no maximum, then it is easy to see that any cofinite subset of $I$ is cofinal, so your desired conclusion will follow.