Impulse response of the inverse system to the backward difference
You're on the right track. Let's call the impulse response of the second system (equalizing the first one) $h_2(n)$. The trick is to assume causality of the second system, i.e. $h_2(n)=0$ for $n&lt;0$. By doing so, the solution for $h_2(n)$ becomes unique.

The total impulse response of the cascaded system is

$$h(n)=h_2(n)-h_2(n-1)\tag{1}$$

and this must equal $\delta(n)$. So we get

$$h_2(0)=1\\
h_2(n)=h_2(n-1),\quad n\ge 1$$

Which gives

$$h_2(1)=h_2(0)=1\\
h_2(2)=h_2(1)=1\\
h_2(3)=h_2(2)=1\\\vdots$$

So $h_2(n)=u(n)$, where $u(n)$ is the discrete-time step function.

Note that you can also find an anti-causal filter $h_2(n)$ satisfying the requirement $h(n)=\delta(n)$. This anti-causal solution is $h_2(n)=-u(-n-1)$. You can easily convince yourself that for this choice of $h_2(n)$ the total impulse response $h(n)$ given by (1) also equals $\delta(n)$.

Note also that all the solutions you gave at the end of the question have the form $h_2(n)=u(n)+c$ where $c$ is some constant. It is indeed true that all these systems are solutions to the problem because the frequency response of the first system has a zero at DC, compensating for any DC component of $h_2(n)$. However, all of these systems with $c\neq 0$ are neither causal nor stable.