Irreducible representation that is not absolutely semisimple

This cannot happen for any $G$, finite or infinite. More generally, it cannot happen for modules over any $k$-algebra which are finite dimensional as $k$-vector spaces.

Lemma Let $k$ be a perfect field and $V$ a finite dimensional $k$ vector space. Let $\bar{W}$ be a $\bar{k}$ subspace of $V \otimes_k \bar{k}$ which is taken to itself by every element of $\mathrm{Gal}(\bar{k}/k)$. Then there is a subspace $W$ of $V$ such that $\bar{W} = W \otimes_k \bar{k}$.

Proof Identify $V$ with $k^n$. We can uniquely write $\bar{W}$ as the row span of a matrix in reduced row echelon form; let the rows of this matrix be $f_1$, $f_2$, ..., $f_r$.

To say this in a more abstract way, fix a basis $e_1$, $e_2$, ..., $e_n$ for $V$; let $i_1$, $i_2$, ..., $i_r$ be the indices for which $\dim \bar{W} \cap (e_{i_a}, e_{i_a +1}, e_{i_a + 2}, ..., e_n) > \dim \bar{W} \cap (e_{i_a +1}, e_{i_a + 2}, ..., e_n)$ and let $f_a$ be the unique vector in $\bar{W}$ of the form $e_{i_a} + \sum_{j > i_a,\ j \neq i_{a+1}, i_{a+2}, ..., i_r} f_{aj} e_j$.

From either the uniqueness reduced row echelon form or the explicit description above, $\bar{W}$ uniquely determines the vectors $f_1$, $f_2$, ..., $f_r$. The condition that $\mathrm{Gal}(\bar{k}/k)$ takes $\bar{W}$ to itself therefore means that $\mathrm{Gal}(\bar{k}/k)$ fixes each $f_a$, so all the vectors $f_a$ lie in $V$. Their span is the required $W$. $\square$.

Now, let $V$ be a semi-simple representation of $G$ as required. Let $\bar{W}$ be the subspace of $\bar{V}$ spanned by all simple subrepresentations of $\bar{V}$. We claim that $\bar{V} = \bar{W}$. Proof: By the previous lemma, there is some $W \subseteq V$ with $\bar{W} = W \otimes_k \bar{k}$. If $W \neq V$, then, since $V$ is semisimple, there is some nonzero subrepresentation $X$ of $V$ with $X \cap W = (0)$. But then $\bar{X} \cap \bar{W} = (0)$, and $\bar{X}$ must contain some simple subrepresentation, a contradiction. $\square$

We have now shown that $\bar{W}$ is spanned by its simple subrepresentations, and we must show that it is semi-simple. We must show that, for any subsrepresentation $A$ of $\bar{W}$, there is a subrepresentation $B$ with $\bar{W} = A \oplus B$. We induct on $\dim \bar{W}/A$; the base case is $\bar{W} = A$ so we can take $B =0$.

If $\bar{W} \supsetneq A$ then there must be some simple subrepresentation $S$ of $\bar{W}$ not lying in $A$. Since $S$ is simple, $S \cap A = (0)$. So $A \oplus S$ is a larger subrepresentation of $\bar{W}$ and, by induction, we can find $B'$ with $\bar{W} = A \oplus S \oplus B'$. Then take $B = S \oplus B'$. $\square$.

This doesn't happen for finite groups. Here's a proof: let $G$ be a finite group and let $L/K$ be a separable field extension. If $V$ is a simple $K[G]$-module, then $\mathrm{rad}(K[G])$ acts trivially on $V$. By Theroem 5.17 in Lam's "A First Course in Noncommutative Rings", we have $\mathrm{rad}(L[G]) = L \otimes_K \mathrm{rad}(K[G])$ (this step uses separability), thus $\mathrm{rad}(L[G])$ acts trivially on $L \otimes_K V$. As $L[G]$ is Artinian, this implies that $L \otimes_K V$ is semisimple, as the module structure factors over the semisimple quotient $L[G]/\mathrm{rad}(L[G])$.

I don't know the answer for general groups. The argument shows at least that $\mathrm{rad}(L[G])$ acts trivially, but that's not enough to conclude that $L \otimes_K V$ is semisimple in general.