For example, when $\sum$ goes from $i=0$ to $k$ on both sides: $$\sum_{i=0}^k n_i b^i \equiv \sum_{i=0}^k n_i \pmod{b-1}$$
Is it possible to divide out the ni on both sides? If not, are there any other ways to simplify this equation that I may be missing?
It is not correct to divide. If it is useful, you might rewrite as $$\sum_{i=0}^kn_i(b^i-1)\equiv 0\pmod{b-1}.$$
It is useful, for example, if you want to prove that your relationship holds for all choices of the $n_i$. That is because one can show that $b^i-1\equiv 0\pmod{b-1}$.
That's just a fancy way of saying that $b-1$ divides $b^n-1$. The result is not hard to see, since we have $$b^n-1=(b-1)(b^{n-1}+b^{n-2}+\cdots+b+1).$$
Remark: It is quite a bit easier to work directly with congruences. Note that $b\equiv 1\pmod{b-1}$. So $b^i\equiv 1\pmod{b-1}$ for all $i$. It follows immediately that $$\sum_0^k n_ib^i\equiv \sum_0^k (n_i)(1)=\sum_0^k n_i.$$
That will give you a generalization of the old-fashioned "casting out nines" process to a base-$b$ system.
We have $b \equiv 1$ modulo $b - 1$ so for any $i$ we have $b^i \equiv 1$ modulo $b - 1$, but this doesn't come from "dividing" the summation, remember that $$\sum x_iy_i$$ means $$x_1y_1 + x_2y_2 + x_3y_3 + \cdots$$ so you can't divide out $\sum x_i$.
Anyway, $b^i \equiv 1$ modulo $b - 1$ is trivially true and you aren't going to get any information about the $n_i$ from it. The only simplification you could make is to combine the summations: $$\sum_{i = 0}^kn_i(b^i - 1) \equiv 0 \pmod{b - 1}$$ but again $b^i - 1 \equiv 0$ modulo $b - 1$ so this won't give you any information about the $n_i$.