I was fiddling around and found that a function of the form $$L_b (x)=\left(\frac{b^{\phi (p^k)}-1}{p^k}\right)^{-1}\left(\frac{x^{\phi (p^k)}-1}{p^k}\right) \mod p^k$$ seems to behave similarly to a discrete logarithm modulo $p^k$ (explained below), assuming $p \nmid x$, $p \nmid b$, and $p^{k+1} \nmid b^{\phi(p^k)}-1$.
For example, if $p^k=9$ and $b=2$, then $L_2(2)=1, L_2(4)=2, L_2(8)=3$.
However, there seems to be some errors, such as (with the same modulus) $L_2 (16)=7$, when its discrete logarithm is actually $4$
To clarify what I meant by having properties similar to a discrete logarithm:
Proposition 1. $L_b (xy)=L_b (x) + L_b (y)$, for $x$ and $y$ coprime to $p^k$.
$\underline{Proof}$: Let $\phi (n)$ be Euler's Totient function. Then
$$(x^{\phi (p^k)}-1)(y^{\phi (p^k)}-1) \equiv 0 \mod p^{2k} \Rightarrow$$
$$(xy)^{\phi (p^k)}-1 \equiv (x^{\phi (p^k)}-1)+(y^{\phi (p^k)}-1) \mod p^{2k} \Rightarrow$$
$$\frac{(xy)^{\phi (p^k)}-1}{p^k} \equiv \frac{x^{\phi (p^k)}-1}{p^k}+\frac{y^{\phi (p^k)}-1}{p^k} \mod p^k \Rightarrow$$
$$L_b (xy) \equiv L_b (x) + L_b (y) \mod p^k$$
Note that in the last step, we multiplied by $\left(\frac{b^{\phi(p^k)}-1}{p^k}\right)^{-1}$. Admittedly, $\phi$ could be replaced with Carmichael's function, and I believe it would still work.
Proposition 2. If $x=b^r$ for residue $r \mod p^k$, then $L_b (x)=r \mod p^k$.
$\underline{Proof}$: We have the following $$L_b (x)=\left(\frac{b^{\phi (p^k)}-1}{p^k}\right)^{-1} \left(\frac{x^{\phi (p^k)}-1}{p^k} \right)=\\\left(\frac{b^{\phi (p^k)}-1}{p^k}\right)^{-1} \left(\frac{(b^r)^{\phi (p^k)}-1}{p^k} \right)=\\ \left(\frac{b^{\phi (p^k)}-1}{p^k}\right)^{-1}\left(\frac{b^{\phi (p^k)}-1}{p^k}+\frac{b^{\phi (p^k)}-1}{p^k}+\dots \frac{b^{\phi (p^k)}-1}{p^k} \right)=\\ 1+1+ \dots +1 \equiv r \mod p^k$$
However, due to the previously mentioned errors, I must have made some sort of mistake.
If someone could look this over, explain the mistake and possibly suggest a better modified function, it would be much appreciated