Is there any chance that this: $$\sum_{s=n}^{m}\binom{m}{s}(x-1)^{m-s} = \sum_{s = n}^{m} \sum_{j=s}^{m}\binom{s - 1}{n - 1}\binom{m-s}{j-s} (x- 1)^{m - j +s - n}$$ is an identity? If so, what is its proof?
I observed this by solving a combinatorics problem where one of the solutions using some observations with combinatorics was $\displaystyle k^m - \sum_{i = 0}^{n - 1}\binom{m}{i}(k - 1)^{m - i}$ and the other was using generating functions and ended up with this formula $\displaystyle\sum_{i = 0}^{m - n} \binom{n + i - 1}{i}(k - 1)^i \cdot k^{m - n - i}$ and somehow they are always equal for any positive integers $m \ge n$ and $k$. By some manipulations, I arrived at the identity above, but I don't know how it's true.
Yes, it is an identity.
Let's substitute $t=x-1$, assume that $m$ is constant and find the genfunc over $n$ for LHS:
$$ \begin{align} \sum\limits_{n=0}^m x^n\sum\limits_{s=n}^m \binom{m}{s} t^{m-s} &=\sum\limits_{s=0}^m \binom{m}{s} t^{m-s} \sum\limits_{n=0}^s x^n \\ &= \sum\limits_{s=0}^m \binom{m}{s} t^{m-s} \frac{1-x^{s+1}}{1-x} \\ &= \frac{1}{1-x} \sum\limits_{s=0}^m \binom{m}{s} t^{m-s} (1-x^{s+1}) \\ &= \boxed{\frac{(1+t)^m - x(x+t)^m}{1-x}} \end{align}$$
For the RHS, let's first simplify it to $$ \begin{align} \sum\limits_{s=n}^m \sum\limits_{j=s}^m \binom{s-1}{n-1} \binom{m-s}{j-s} t^{m-j+s-n} &= \sum\limits_{s=n}^m \binom{s-1}{n-1} \sum\limits_{j=0}^{m-s} \binom{m-s}{j} t^{s-n} t^{m-s-j} \\ &= \sum\limits_{s=n}^m \binom{s-1}{n-1}t^{s-n} (1+t)^{m-s}. \end{align}$$
Now let's also find the genfunc over $n$:
$$ \begin{align} \sum\limits_{n=0}^m x^n\sum\limits_{s=n}^m \binom{s-1}{n-1}t^{s-n} (1+t)^{m-s} &= \sum\limits_{s=0}^m (1+t)^{m-s} \sum\limits_{n=0}^s \binom{s-1}{n-1} t^{s-n} x^n \\ &= (1+t)^m + \sum\limits_{s=1}^m (1+t)^{m-s} x(t+x)^{s-1} \\ &= (1+t)^m + x(1+t)^{m-1}\sum\limits_{s=0}^{m-1} \frac{(t+x)^{s}}{(1+t)^s}. \end{align}$$
This further rewrites as $$\begin{align} &= (1+t)^m + x(1+t)^{m-1} \left( \frac{1-\frac{(t+x)^m}{(1+t)^m}}{1-\frac{t+x}{1+t}} \right) \\ &= (1+t)^m + x(1+t)^{m-1} \frac{(1+t)^m-(t+x)^m}{(1+t)^m - (t+x)(1+t)^{m-1}} \\ &=(1+t)^m + \frac{x(1+t)^m - x(t+x)^m}{1+t-(t+x)} \\ &=(1+t)^m + \frac{x(1+t)^m - x(t+x)^m}{1-x} = \boxed{\frac{(1+t)^m-x(x+t)^m}{1-x}} \end{align}$$
P.S. Of course, another way would be to simply group together coefficients near $t^{m-s}$ in RHS:
$$ \sum\limits_{s=n}^m \sum\limits_{j=s}^m \binom{s-1}{n-1} \binom{m-s}{j-s} t^{m-j+s-n} = \sum\limits_{r=n}^m t^{m-r} \sum\limits_{s=n}^m \binom{s-1}{n-1} \binom{m-s}{r-n}. $$
Now we just need to show that $$\begin{align} \sum\limits_{s=n}^m \binom{s-1}{n-1} \binom{m-s}{r-n} = \binom{m}{r}, \end{align}$$ which might or might not be easier to do directly.