Okay, this is getting too complicated for comments.
Let $\{S_i; (f_{ji})\}_{i\in I}$ be the inverse system. I take $I$ to be a directed partially ordered set, $S_i$ simple for each $i$, and whenever $i\leq j$, $f_{ji}$ is an onto homomorphism $f_{ji}\colon S_j\to S_i$.
Since an onto homomorphism with domain a simple group must be either the trivial map or an isomorphism, we see that for every $i$ and $j$ in $I$, with $i\leq j$, either $S_i\cong S_j$ or else $S_i=\{1\}$.
In particular, let $i$ and $j$ be arbitrary elements of $I$. Then there exists $k$ such that $i\leq k$ and $j\leq k$. Hence if $S_i$ and $S_j$ are both nontrivial, then $S_k\cong S_i$, $S_k\cong S_j$, and hence $S_i\cong S_j$. So any two nontrivial groups in our family are necessarily isomorphic.
If all groups are trivial, then the inverse limit is trivial. If not every group is trivial, then let $S$ be a simple group such that every nontrivial $S_i$ is isomorphic to $S$. I claim that in this case, the inverse limit is isomorphic to $S$.
The inverse limit is given by the subgroup of $\prod_{i\in I}S_i$ of all elements $(x_i)$ such that if $i\leq j$, the $f_{ji}(x_j) = x_i$. We may omit any component where $S_i=\{1\}$, and so we may assume that every $S_i$ is isomorphic to $S$ and that all structural morphisms are isomorphisms. But then I claim that $\lim\limits_{\leftarrow}S_i\cong S$. Indeed, for any $i$ and any $x_i\in S_i$, for every $j\geq i$ there is a unique $x_j\in S_j$ such that $f_{ji}(x_j) = x_i$. if $j\leq k$, then $f_{ji}(x_j) = x_i = f_{ki}(x_k) = f_{ji}(f_{kj}(x_k))$, so the uniqueness guarantees that $f_{kj}(x_k) = x_j$. We can complete the terms "downstream" as well: given any $j$, find $k$ with $i,j\leq k$; then take $x_j = f_{kj}(x_k)$, where $x_k$ was defined as above. So we get a consistent family that has $x_i$ in the $i$th coordinate. Hence $\pi_i$, the projection onto the $i$th coordinate, is onto.
I claim that it is one-to-one; indeed, if $\pi_i(x_i) = 1$, then $x_i=1$. Now let $j\in I$ be arbitary. Then there exists $k\in I$ with $i,j\leq k$; since $f_{ki}$ is an isomorphism and $f_{ki}(x_k) = x_i = 1$, then $x_k=1$; hence $f_{kj}(x_k) = x_j = 1$ as well. Thus, $(x_i)= 1$, so $\pi_i$ is one-to-one. Since $\pi_i$ is one-to-one and onto, $\pi_i$ is an isomorphism, so $\lim\limits_{\leftarrow}S_i \cong S_i \cong S$, as desired.