Solving Difference Equation Using Z-Transform
Your approach is correct, but the problem is that the given response is not the response to a unit step, but to a scaled step. That's why you get \$\alpha y[-1]=-11/2\$ and \$1-\alpha y[-1]=8\$, which is incompatible. If you assume a scaled step \$ku[n]\$, you end up with the following \$\mathcal{Z}\$-transform of the output signal:

$$Y(z)=\frac{k-\alpha y[-1]+\alpha y[-1]z^{-1}}{(1-z^{-1})(1+\alpha z^{-1})}\tag{1}$$

Comparing \$(1)\$ to the \$\mathcal{Z}\$-transform of the given response you get

$$k-\alpha y[-1]=8,\quad \alpha y[-1]=-\frac{11}{2},\quad\alpha=-\frac12$$

from which you obtain

$$y[-1]=11\quad\text{and}\quad k=\frac{5}{2}$$

The values of \$\alpha\$ and \$y[-1]\$ do not change, but now the result is compatible with the given response.