Fourier transform of discrete time unit step function
The DFT of a unit step response is
$$U(\omega) = \frac{1}{1 - e^{-j \omega}} + \pi \delta(\omega)$$
Applying the shift property as you did will give:
$$\mathcal{F}(u[n] - u[n-1]) = U(\omega) - U(\omega)e^{-j \omega} = \frac{1}{1 - e^{-j \omega}} + \pi \delta(\omega) - [\frac{1}{1 - e^{-j \omega}} + \pi \delta(\omega)]e^{-j \omega}$$
that is
$$\mathcal{F}(u[n] - u[n-1]) =  \frac{1- e^{-j \omega}}{1 - e^{-j \omega}} +\pi\delta(\omega)( 1 - e^{-j\omega}) = 1+\pi\delta(\omega)( 1 - e^{-j\omega})$$
The second term is always zero because for $\omega = 0$, $1 - e^{-j\omega} = 0 $ and it's zero on any other point, so you get,
$$\mathcal{F}(u[n] - u[n-1]) = 1 = \mathcal{F}(\delta[n])$$