Impulse Response with non-delayed feedback
Take the z-transform of your equation to get $y(z) = x(z) h(z) - y(z) h(z)$ which you can simplify to $$ y(z) \left( 1 + h(z) \right) = x(z) h(z) $$ and therefore get the transfer function 
$$ H(z) := \frac{y(z)}{x(z)} = \frac{h(z)}{1+h(z)} $$

Now plug in the z-transform $h(z)$ of your response $h(n)$ and perform an inverse z-transform to get the total impulse response $H(n)$.

An approach that is more basic and does not require z-transforms works for your particular example by solving the recursion equation explicitly. The inner system with the response $h(n) = u(n) \left( \frac{1}{2}\right)^n$ can be written with the recursive difference equation $y'(n) = x'(n) + \frac{1}{2} y'(n-1)$ where $x'$ and $y'$ are the output and input of the subsystem. Then we can write the full system in terms of the following equations:

$$ y(n) = y'(n)$$
$$ x'(n) = x(n) - y(n)$$
$$ y'(n) = x'(n) + \frac{1}{2} y'(n-1) $$

Eliminating $x'$ and $y'$ results in

$$ y(n) = x(n) - y(n) + \frac{1}{2} y(n-1) $$

which simplifies to

$$y(n)=\frac{1}{2}x(n)+\frac{1}{4}y(n-1)$$

This system has the impulse response $$H(n) = \frac{1}{2} u(n) \left(\frac{1}{4}\right)^n$$