How to find the sum of $1+(1+r)s+(1+r+r^2)s^2+\dots$?

Your sum can be written as $$\sum_{n=0}^{\infty}\left(\sum_{k=0}^{n}r^k\right)s^n=\sum_{n=0}^{\infty}\frac{1-r^{n+1}}{1-r}s^n=\frac{1}{1-r}\sum_{n=0}^{\infty}s^n- \frac{r}{1-r}\sum_{n=0}^{\infty}(rs)^n.$$ Can you take it from here?

P.S. Here we are assuming that $|s|<1$, $|rs|<1$ and $r\not=1$. What happens when $r=1$?

If you multiply your series by $r-1$, you will get$$(r-1)+(r^2-1)s+(r^3-1)s^2+\cdots,\tag1$$which is the sum of$$r+r^2s+r^3s^2+\cdots$$with$$-1-s-s^2-\cdots$$The sum of the first series is $\frac r{1-rs}$, whereas the sum of the second one is $-\frac1{1-s}$. Therefore,$$(1)=\frac1{r-1}\left(\frac r{1-rs}-\frac1{1-s}\right)=\frac1{(1-s)(1-rs)}.$$

You expanded the brackets, but did not actually group:

$$ 1 + (1+r)s + (1+r+r^2)s^2 + \dotsb = 1 + (s+rs \color{red}{{} + \dotsb}) + (s^2+rs^2+r^2s^2\color{red}{{} + \dotsb}). $$

Here is the way to group: \begin{align*} & 1+(1+r)s+(1+r+r^2)s^2+\dotsb \\ &= 1+(\color{red}{s}+\color{green}{rs})+(\color{red}{s^2}+\color{green}{rs^2}+\color{blue}{r^2s^2})+\dotsb \\ &= (1+\color{red}{s}+\color{red}{s^2}+\dotsb)+(\color{green}{rs}+\color{green}{rs^2}+\dotsb)+(\color{blue}{r^2s^2}+r^2s^3+\dotsb) \\ &= \frac{1}{1-s}+\frac{rs}{1-s}+\frac{r^2s^2}{1-s}+\dotsb \\ &= \frac{1}{1-s}(1+rs+r^2s^2+\dotsb) \end{align*} Can you finish?