Is the following system stable and causal?
You are on the right track, but you are confused.
First, regarding the calculations you could simply do 
$$y[n]=2\left(\frac{1}{2}\right)^n(u[n]+\delta[n+1])=2\left(\frac{1}{2}\right)^nu[n]+4\delta[n+1]$$
which gives you 

$$Y(z)=\frac{2z}{z-\frac{1}{2}}+4z,\,|z|&gt;\frac{1}{2}$$
The impuse resposne is the first difference of the step response:

$$\begin{align}
h[n]&amp;=y[n]-y[n-1]\\
&amp;=2\left(\frac{1}{2}\right)^nu[n]+4\delta[n+1]-4\left(\frac{1}{2}\right)^nu[n]\\
&amp;=4\delta[n+1]-2\left(\frac{1}{2}\right)^nu[n]
\end{align}$$
Obviously, $h[-1]=4$ and the system is not causal.
But here you are mixing up finding the inverse $z$-transform of anti-casual signals with system being non-causal. ROC is $|z|&gt;\frac{1}{2}$ and the system is non-causal full stop.