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We know that $Z^{-1}\left(\frac{z}{z-a}\right) = a^nu[n]$ if $|z| > |a|$.
In addition, $Z^{-1}\left(\frac{1}{z-a}\right) = a^{n-1}u[n-1]$ if $|z| > |a|$. This is the delayed version of the first one.
In acausal case, $Z^{-1}\left(\frac{z}{z-a}\right) = -a^nu[-n-1]$ if $|z| < |a|$.
But, I am confused about $Z^{-1}\left(\frac{1}{z-a}\right)$ if $|z| < |a|$. Is $Z^{-1}\left(\frac{1}{z-a}\right)$ equal to $-a^{n-1}u[-n-2]$ or $-a^{n-1}u[-n]$ or something else.
Regardless whether a sequence is causal or not, a multiplication with $z^{-1}$ is always a delay by one sample. I.e., if $X(z)$ is the $\mathcal{Z}$-transform of $x[n]$, then $z^{-1}X(z)$ is the $\mathcal{Z}$-transform of $x[n-1]$.
So if we know that for $|z|<|a|$
$$\mathcal{Z}^{-1}\left\{\frac{z}{z-a}\right\}=-a^nu[-n-1]$$
then we must have
$$\mathcal{Z}^{-1}\left\{\frac{1}{z-a}\right\}=-a^{n-1}u[-(n-1)-1]=-a^{n-1}u[-n]$$
