A zeta function for the Klein Four group? Let $G$ be a finite group, $S \subset G$ a generating set, $|g|:=|g|_S=$ word-length with respect to $S$. Let $\phi(g,h)=|g|+|h|-|gh| \ge 0$ be the "defect-function" of $S$. The set $\mathbb{Z}\times G$ builds a group for the following operation:
$$(a,g) \oplus (b,h) = (a+b+\phi(g,h),gh)$$
On $\mathbb{N}\times G$ is the "norm": $|(a,g)| := |a|+|g|$ additive, which means that $|a \oplus b| = |a|+|b|$. Define the multiplication with $n \in \mathbb{N_0}$ to be:
$$ n \cdot a := a \oplus a \oplus \cdots \oplus a$$
(if $n=0$ then $n \cdot a := (0,1) \in \mathbb{Z} \times G$).
A word $w := w_{n-1} w_{n-2} \cdots w_0$ is mapped to an element of $\mathbb{Z} \times G$ as follows:
$$\zeta(w) := \oplus_{i=0}^{n-1} (m^i \cdot (0,w_i))$$
where $m := \min_{g,h\in G, \phi(g,h) \neq 0} \phi(g,h)$.
We let $|w|:=|\zeta(w)|$ and $w_1 \oplus w_2:=\zeta(w_1)\oplus \zeta(w_2)$
Then we have $|w_1 \oplus w_2| = |w_1|+|w_2|$.
For instance for the Klein four group $\{0,a,b,c=a+b\}$ generated by $S:=\{a,b\}$, we get sorting the words $w$ by their word-length:
$$0,a,b,c,a0,aa,ab,ac,b0,ba,bb,bc,c0,ca,cb,cc,a00,a0a,a0b,a0c$$
corresponding to the following $\mathbb{Z}\times K_4$ elements $\zeta(w)$:
$$(0,0),(0,a),(0,b),(0,c),(2,0),(2,a),(2,b),(2,c),(2,0),(2,a),(2,b),(2,c),(4,0),(4,a),(4,b),(4,c),(4,0),(4,a),(4,b),(4,c)$$
corresponding to the the following "norms" of words $|w| = |\zeta(w)|$:
$$0,1,1,2,2,3,3,4,2,3,3,4,4,5,5,6,4,5,5,6$$
Let $a_n, n\ge 0$ be the sequence of numbers generated by the Klein four group.
1) Is 
$$\sum_{n=1}^\infty \frac{1}{a_n^s} = \sum_{n=1}^\infty \frac{n+1}{n^s} = \zeta(s-1) + \zeta(s)$$
where $\zeta$ denotes the Riemann zeta function?
I have checked this with SAGE math up to a certain degree and it seems plausible, however I have no idea how to prove it.
2) Is every $a_n$ the product of primes $p=a_k$ for some $k\le n$?
3) Let $\pi_{K_4}(n) = |\{ k : \text{$a_k$ is prime, $k \le n$}\}|$ be the prime counting function of the sequence.  What is the approximate relationship to the usual prime counting function $\pi(n)$?
 A: Consider a word $w=w_0w_1\cdots w_{n-1}$ with each $w_i\in\{0,a,b,c\}$.
Following your notation,
$$\lvert w\rvert=\lvert\zeta(w)\rvert=\left\lvert\bigoplus_{i=0}^\infty m^i\cdot(0,w_i)\right\rvert=\sum_{i=0}^{n-1}\lvert 2^i\cdot(0,w_i)\rvert=\sum_{i=0}^{n-1} 2^i\lvert w_i\rvert$$
where $\lvert0\rvert=0$, $\lvert a\rvert=1$, $\lvert b\rvert=1$, $\lvert c\rvert=2$.
We now consider the generating function
$$F_n(x)=\sum_wx^{\lvert w\rvert}$$
where the sum is over all words $w=w_0w_1\cdots w_{n-1}$.
Then
\begin{align*}
F_n(x)&=\sum_{w_0}\sum_{w_1}\cdots\sum_{w_{n-1}}x^{|w_0|+2|w_1|+\cdots+2^{n-1}|w_{n-1}|}\\
&=\left(\sum_{w_0}x^{|w_0|}\right)\left(\sum_{w_1}x^{2|w_1|}\right)\cdots\left(\sum_{w_{n-1}}x^{2^{n-1}|w_{n-1}|}\right)\\
&=(1+2x+x^2)(1+2x^2+x^4)(1+2x^4+x^8)\cdots(1+2x^{2^{n-1}}+x^{2^n})\\
&=(1+x)^2(1+x^2)^2(1+x^4)^2\cdots(1+x^{2^{n-1}})^2\\
&=(1+x+x^2+\cdots+x^{2^n-1})^2\\
&=1+2x+\cdots+(2^n-1)x^{2^n-2}+2^nx^{2^n-1}+(2^n-1)x^{2^n}+\cdots+x^{2^{n+1}-2}.
\end{align*}

What this shows is that among the first $4^n$ terms of the sequence:
\begin{align*}
&0\text{ appears }1\text{ times},\\
&1\text{ appears }2\text{ times},\\
&2\text{ appears }3\text{ times},\\
&\cdots\\
&2^n-2\text{ appears }2^n-1\text{ times},\\
&2^n-1\text{ appears }2^n\text{ times},\\
&2^n\text{ appears }2^n-1\text{ times},\\
&\cdots\\
&2^{n+1}-4\text{ appears }3\text{ times},\\
&2^{n+1}-3\text{ appears }2\text{ times},\\
&2^{n+1}-2\text{ appears }1\text{ times}.\\
\end{align*}
This resolves question 1.
Since this seems to be purely a combinatorics question, I would not expect the second and third questions to have particularly interesting answers.

I will remark that
$$\pi_{K_4}(4^n)=\sum_{p\leq2^{n+1}-2}(2^n-\lvert p-2^n+1\rvert)p$$
so you could use some analytic number theory to get asymptotics for $\pi_{K_4}(n)$ if you wanted.
