Is the $k$-algebra $k[G]$ semisimple?

I don't know where to find a reference for the claim I made in the comments, so here is a proof. It feels annoyingly complicated to me; I don't know if a substantially simpler proof is possible.

Theorem (Maschke): Let $k$ be a field and let $G$ be a finite group. Then $k[G]$ is semisimple iff the characteristic of $k$ does not divide $|G|$.

Proof. First suppose that the characteristic of $k$ divides $|G|$; call it $p$. We will use the fact that a finite-dimensional $k$-algebra is semisimple iff its Jacobson radical vanishes (see e.g. this blog post). Our goal will be to exhibit an element of $k[G]$ in its Jacobson radical. Let

$$s = \sum_{g \in G} g \in k[G].$$

Note that $gs = s$ for all $g \in G$ and that $s^2 = |G| s = 0$; in particular, $s$ is nilpotent, but moreover the left ideal generated by $s$, which consists of scalar multiples of $s$, consists of nilpotent elements. It follows that $s$ is a nonzero element of the Jacobson radical of $k[G]$, and hence that $k[G]$ is not semisimple.

Now suppose that the characteristic of $k$ does not divide $|G|$; call it $p$. Recall that if $A$ is any finite-dimensional $k$-algebra, then there is a canonical linear functional on $A$ called the trace

$$\text{tr}(a) = \text{tr}(L_a), L_a : A \ni x \mapsto ax \in A.$$

That is, $\text{tr}(a)$ is the trace of $a$ acting on $A$ by left multiplication. The trace induces a symmetric bilinear form on $A$ called the trace form $\langle a, b \rangle = \text{tr}(ab)$. Recall that a bilinear form on a vector space is nondegenerate if $a \neq 0$ implies that there exists some $b$ such that $\langle a, b \rangle \neq 0$.

Lemma: If the trace form on $A$ is nondegenerate, then $A$ is semisimple.

Proof. Let $j \in J(A)$ be an element of the Jacobson radical of $A$. On a finite-dimensional $k$-algebra, $J(A)$ can be characterized as the largest nilpotent right ideal of $A$. Hence if $b \in A$, then $jb \in J(A)$ is nilpotent. By passing to an algebraic closure of $k$ and upper-triangularizing the action of $jb$ on $A$ by left multiplication we see that the eigenvalues of $jb$ are all zero, so $\text{tr}(jb) = \langle j, b \rangle = 0$. It follows that if $j \neq 0$ then the trace form is degenerate. $\Box$

The trace on $k[G]$ takes a very simple form: $\text{tr}(1) = |G|$, and $\text{tr}(g) = 0$ for all non-identity elements $g$. By assumption, $|G| \neq 0$ in $k$, so it follows that if $a = \sum a_g g, a_g \neq 0$, then

$$\text{tr}(g^{-1} a) = |G| a_g \neq 0$$

and hence the trace form is nondegenerate. The conclusion follows. $\Box$

Your mistake is assuming that $k_1^p + k_2^p \neq 0$.

Hint: As $G$ is cyclic, say with generator $g$, we have $$k[G] = k[g]/(g^p - 1).$$ Show that the map $$k[x]/x^p \to k[G]$$ defined by $x \mapsto g - 1$ is an isomorphism. Then it shouldn't be hard to see that $k[x]/x^p$ is not semisimple.

For the very reason you indicate, you can see that $k[G]$ contains a non-zero nilpotent element: let $g$ be a generator of $G$: the $(1_{G}-g)^{p} = 1_{G} - g^{p} = 0.$ In particular, if $M$ is an irreducible $k[G]$-module, then $0$ is the only eigenvalue of $1_{G}-g,$ and $1$ is the only eigenvalue of $g$ on $M$. Now as $G$ is Abelian, $M(1_{G}-g)$ is a $k[G]$-submodule of $M$, so is $\{0\}$, as $1_{G}-g$ is not invertible.Since $g$ generates $G,$ we see that $M$ must be the trivial module. Hence $k[G]$ is not semisimple, since the regular module is certainly not a direct sum of trivial modules.