A $p$-adic sum of reciprocals of powers

Let $p$ be a prime number and $k\geq 2$ an even integer. Consider the following $p$-adic integer: $$S_{p,k} := \lim_{r\to+\infty} \sum_{a=1}^{p^r} \big(\frac{p^r}{a}\big)^k$$ Convergence is easy to see. This can also be written as $$S_{p,k} = \frac{1}{2} \sum_{0\neq x \in B_p} x^{-k}$$ where \begin{aligned} B_p &:= \big\{\frac{a}{p^r}\; : \; a\in\mathbb{Z},\; r\in\mathbb{N},\; |a|_{\infty} \leq p^r\big\}\\ &=\{x\in\mathbb{Q} : |x|_v \leq 1 \text{ at all places } v\neq p\}\\ \end{aligned} (Here, $|\cdot|_\infty$ is the usual real absolute value, and “places” [of $\mathbb{Q}$] mean either a prime $\ell$ or the symbol $\infty$.) The point of this last presentation is that it suggests the analogy with $B_\infty = \mathbb{Z}$, where $S_{\infty,k} = \sum_{n=1}^{+\infty} n^{-k}$ is the usual Riemann zeta function (at even integers): in a certain sense, the $S_{p,k}$ are a $p$-adic analogue of the $\zeta(k)$ (though probably not the most intelligent or satisfactory analogue).

Question(s): Have these $S_{p,k}$ appeared in the literature? Do they have a name? Do they satisfy some known relations? (E.g., can $S_{p,4}$ be expressed in function of $S_{p,2}^2$? Experimentally, the ratio $S_{p,4}/S_{p,2}^2$ does not appear to be rational, but maybe something else can be said.)

For what it's worth, Wolstenholme's theorem $\sum_{a=1}^{p-1}\frac{1}{a^2}\equiv 0\pmod{p}$ tells us that $S_{p,2} \equiv 1\pmod{p^4}$. There seem to be many generalizations in various directions (see, e.g., here) involving Bernoulli numbers, Kummer congruences, or the Kubota-Leopoldt $p$-adic zeta function, but I couldn't find a direct connection with the above quantity.

As far as I know this quantity does not have a name. There is a mention of a general family of $p$-adic limits, of which $S_{k,p}$ is a special case, on p. 31 of [1].
The quantity $S_{k,p}$ can be expressed as an infinite series involving the Kubota-Leopoldt $p$-adic zeta function $\zeta_p(n):=L_p(n,\omega_p^{1-n})$: $$S_{p,k}=1-\sum_{n\geq 1}{-k\choose n}\left(1-p^{-(n+k)}\right)^{-1}\zeta_p(n+k).$$ This can be derived from the main formula of [2], and something much more general appears in [1].
The even $p$-adic zeta values vanish, and the odd ones are expected to be algebraically independent. We can define a sort of formal version of $S_{k,p}$ by $$s_{k} := 1-\sum_{\substack{n\geq 1\\n+k\text{ odd}}}{-k\choose n} z_{n+k} \in \mathbb{Q}[[z_3,z_5,\ldots]].$$ One can check that the elements $s_k$ are algebraically independent in $\mathbb{Q}[[z_3,z_5,\ldots]]$ (since they are homogeneous of degree $1$, it suffices to check they are linearly independent, and this is true because the coefficients grow like polynomials of different degrees). So at least there aren't any "obvious" algebraic relations among the $S_{k,p}$ coming from the expression in terms of $\zeta_p$. My guess is that the $S_{k,p}$ are algebraically independent.
[2] Lawrence C. Washington, MR 1611077 $p$-adic $L$-functions and sums of powers, J. Number Theory 69 (1998), no. 1, 50--61.