In this particular problem you can simplify the system by letting $u=z-1$, so that your decomposition becomes
$$\frac{A_1}{u^3}+\frac{B_1}{u^2}+\frac{C_1}u+\frac{A_2}{(u+2)^3}+\frac{B_2}{(u+2)^2}+\frac{C_2}{(u+2)}\;,$$
leading to the equation
$$A_1(u+2)^3+B_1u(u+2)^3+C_1u^2(u+2)^3+A_2u^3+B_2u^3(u+2)+C_2u^3(u+2)^2=1\;.$$
Now the terms can be multiplied out mentally:
$$\begin{array}{ccc}
&&&&A_1u^3&+&6A_1u^2&+&12A_1u&+&8A_1\\
&&B_1u^4&+&6B_1u^3&+&12B_1u^2&+&8B_1u\\
C_1u^5&+&6C_1u^4&+&12C_1u^3&+&8C_1u^2\\
&&&&A_2u^3\\
&&B_2u^4&+&2B_2u^3\\
C_2u^5&+&4C_2u^4&+&4C_2u^3
\end{array}$$
The system of equations is then
$$\left\{\begin{align*}
C_1+C_2&=0\\
B_1+6C_1+B_2+4C_2&=0\\
A_1+6B_1+12C_1+A_2+2B_2+4C_2&=0\\
6A_1+12B_1+8C_1&=0\\
12A_1+8B_1&=0\\
8A_1&=1\;.
\end{align*}\right.$$
This is quickly solved from the bottom up for $A_1,B_1$, and $C_1$:
$$\begin{align*}
&A_1=\frac18\;;\\\\
&2B_1=-3A_1=-\frac38,B_1=-\frac3{16}\;;\\\\
&4C_1=-6B_1-3A_1=\frac98-\frac38=\frac34,C_1=\frac3{16}\;.
\end{align*}$$
The remaining three are quickly found from the top down:
$$\begin{align*}
&C_2=-C_1=-\frac3{16}\;;\\\\
&B_2=-B_1-6C_1-4C_2=\frac3{16}-2C_1=\frac3{16}-\frac6{16}=-\frac3{16}\;;\\\\
&A_2=-A_1-(6B_1+12C_1+2B_2+4C_2)=-\frac18\;,
\end{align*}$$
since the quantity in parentheses on the last line is easily seen to be $0$. In short, $A_1=\frac18$, $A_2=-\frac18$, $B_1=B_2=C_2=-\frac3{16}$, and $C_1=\frac3{16}$. There’s still a fair bit of computation, but it’s a lot less than with the original variable, and it’s simple enough that I can do it about as fast as I can write.