Analytic continuation of Riemann zeta $\zeta(s)$ from the complex $\mathbb{C}$ to quaternion $\mathbb{H}$? One way to define  Riemann zeta function is by the analytic continuation of
$$\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} = \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \cdots$$
for the domain $Re(s)>1$ to the full complex plane in $\mathbb{C}$.
Thus, Riemann zeta function is defined for $s \in \mathbb{C}$ and $\zeta(s)  \in  \mathbb{C}$

My question is that do we gain anything new to do analytic continuation of Riemann zeta function such that a "modified Riemann zeta function" so
$s  \in  \mathbb{H}$ is in quaternion? and $\zeta(s)  \in  \mathbb{H}?$

Does this lead to any interesting result in the math literature?
Edit: more precisely, according to the comment, we seek for an analytic continuation of $\zeta(s)$ from the complex $\mathbb{C}/\{1\}$ to quaternion $\mathbb{H}/\{1\}$?

$\Bbb{H}$ is just a sub-algebra of $M_2(\Bbb{C})$.
For $A \in M_n(\Bbb{C})$ use the Jordan normal form to obtain $A = P J P^{-1} = P (D+N)P^{-1}$ where $D$ is diagonal and $DN=ND$ and $N^n = 0$. Let $f(s) = (s-1)\zeta(s)= \sum_{k=0}^\infty c_k s^k$ which is entire then $$P^{-1} f(A)P  =f(D+N)=\sum_{k=0}^\infty c_k (D+N)^k =\sum_{k=0}^\infty c_k \sum_{l=0}^{n-1} {k \choose l} D^{k-l}N^l= \sum_{k=0}^{n-1} \frac{N^k}{k!} f^{(k)}(D)$$
Note the obtained function $A \mapsto f(A)$ doesn't depend on the basepoint $s_0 = 0$ we chose to expand $f(s)$ in power series.
It is not hard to convince that something similar happens with a meromorphic function such as $\zeta(s)$ obtaining $$\zeta(A) = P \zeta(D+N)P^{-1}=   P \sum_{k=0}^{n-1} \frac{N^k}{k!} \zeta^{(k)}(D)P^{-1}$$
where $\zeta^{(k)}(D)$ is the matrix of $k$-th derivatives $$\zeta^{(k)}(D) = \pmatrix{\zeta^{(k)}(D_{11}) &  & \\  &  \zeta^{(k)}(D_{22}) & \\ & & \ldots}$$
If $q \in \Bbb{H}\subset M_2(\Bbb{C})$ then $q q^* = q^* q = N(q) I$ so that $$q = P DP^{-1}, \qquad \zeta(q)  =P \zeta(D) P^{-1}$$
