obtaning discrete-time system impulse response $h[n]$ from step response $s[n]$
Considering an LTI system, its impulse response $h[n]$ and the step response $s[n]$ are related by $$h[n] = s[n]-s[n-1]$$ and $$s[n] = u[n] \star h[n] = \sum_{k=-\infty}^{n} h[k] = \sum_{k=0}^{\infty} h[n-k] $$ 

Given your definition of $s[n]$ therefore its $h[n]$ is
$$h[n] = \left[5 - 4 \left( \frac{4}{5} \right)^n \right] u[n] -\left[5 - 4 \left( \frac{4}{5} \right)^{n-1} \right] u[n-1]$$

After simplifications it would become:
$$h[n] = 5\delta[n] + 5 \left( \frac{4}{5} \right)^{n} u[n-1] - 4 \left( \frac{4}{5} \right)^{n} u[n]$$ further simplifies to

$$ \boxed{ h[n] = 5 \left( \frac{4}{5} \right)^{n} u[n] - 4 \left( \frac{4}{5} \right)^{n} u[n] = \left( \frac{4}{5} \right)^{n} u[n] }$$

Looking at the signal flow graph and using Z-transform variables you may achieve the same result by considering:

$$Y(z) = \frac{4}{5} Y(z) z^{-1} + X(z)$$
$$Y(z)( 1 - \frac{4}{5} z^{-1}) = X(z)$$
$$ H(z) = \frac{Y(z)}{X(z)} = \frac{1}{1 - \frac{4}{5} z^{-1}} $$

From which the inverse Z-transform (with the causal ROC) yields:
$$ \boxed{ h[n] = \left( \frac{4}{5} \right)^n u[n] }$$ which is the same result obtained from the time-domain approach.