To prove everything in one shot, try arguing as follows:

$S_{n+1} = \sum_{i=1}^{n+1}a_i = (a_{n+1} + \cdots + a_2) + a_1 = ((a_n+r) + \cdots + (a_1+r))+ a = (a_n+\cdots +a_1) + nr + a = S_n + nr + a$.

Now you can apply the inductive hypothesis to $S_n$.

To prove everything in one shot, try arguing as follows:

$S_{n+1} = \sum_{i=1}^{n+1}a_i = (a_{n+1} + \cdots + a_2) + a_1 = ((a_n+r) + \cdots + (a_1+r))+ a = (a_n+\cdots +a_1) + nr + a = S_n + nr + a$.

Now you can apply the inductive hypothesis $S_n = an + \frac{n(n-1)}{2}r$, after which you arrive at the desired expression for $S_{n+1}$ with little extra effort.

The reason you ran into an algebraic problem is you attempted to show $S_k + (a+r) = S_{k+1}$ when instead what you needed was that $S_k + (a+kr) = S_{k+1}$.