Stability of system with poles inside unit circle - conflict with differential equation
What you are missing is that this is about a discrete-time system, because we're talking about poles and zeros in the complex $z$-plane and about poles inside or outside the unit circle. So there is no differential equation, but there is a difference equation:

$$y[n]=\frac12y[n-1]+x[n]\tag{1}$$

The corresponding impulse response is

$$h[n]=\left(\frac12\right)^nu[n]\tag{2}$$

where $u[n]$ is the unit step sequence.

A continuous-time system with a pole at $s=\frac12$ would indeed be unstable, because for stability, all poles of a continuous-time system must be in the left half plane. In discrete time this corresponds to the requirement of all poles being inside the unit circle $|z|=1$.