Is there a way to obtain the impulse response of a discrete system by just knowing it's response to the discrete unit step function?

In continuous time it was possible;

$$u(t){\longrightarrow} \boxed{\quad\textrm{system}\quad} {\longrightarrow} y(t)\implies \delta(t)=\frac{du(t)}{dt}{\longrightarrow}\boxed{\quad\textrm{system}\quad}{\longrightarrow} \frac{dy(t)}{dt}=h(t)$$

Does the same apply for discrete time system i.e.

$$\delta[t]=\frac{du[t]}{dt} \quad\textrm{where:}\begin{cases} \delta[t] &\textrm{is the discrete time delta}\\ u[t] & \textrm{is the discrete time unit step function}\end{cases}$$

Is there a way to obtain the impulse response of a discrete system by just knowing the response of the discrete unit step?

Answer

A simpler version of Phonon's answer is as follows.

Suppose that $y$ denotes the response of the system to the unit step function. Then, as discussed in this answer, in general, $y$ is the sum of scaled and time-delayed copies of the impulse response, and in this particular case, no scaling is required; only time delays. Thus,

$$\begin{align} y[0] &= h[0]\\ y[1] &= h[1] + h[0]\\ y[2] &= h[2] + h[1] + h[0]\\ y[3] &= h[3] + h[2] + h[1] + h[0]\\ \vdots~ &= ~\vdots \end{align}$$ where each column on the right is a (unscaled and) time-delayed impulse response. Thus, we easily get that $$\begin{align} h[0] &= y[0]\\ h[1] &= y[1]-y[0]\\ h[2] &= y[2]-y[1]\\ \vdots~ &= ~\vdots\\ h[n] &~= y[n] - y[n-1]\\ \vdots~ &= ~\vdots \end{align}$$ with nary a mention of filters, inverses, convolutions, integration, operators and the like, just simple consequences of the definition of linear time-invariant system.