How about this:

Note that $$\begin {align} {n \choose r} &= \frac {n(n-1)(n-2)...(n-r+1)} {1\cdot2\cdot3\cdot...(r-1)r} \\ &=\frac nr \cdot \frac {(n-1)(n-2)...(n-r+1)} {1\cdot2\cdot3\cdot...(r-1)} \\ &= \frac nr {n-1 \choose r-1} \\ \Rightarrow {n-1\choose r-1}&=\frac rn {n\choose r} \end{align} $$

and also $$\begin {align} {n \choose r} &= \frac {n(n-1)(n-2)...(n-r+1)} {1\cdot2\cdot3\cdot...(r-1)r} \\ &= \frac n{n-r} \cdot \frac {(n-1)(n-2)...(n-r+1)(n-r)} {1\cdot2\cdot3\cdot...(r-1)r}\\ &= \frac n{n-r} {n-1 \choose r}\\ \Rightarrow {n-1 \choose r}&=\frac {n-r}n {n\choose r} \end{align}$$

So $$\begin {align} {n-1 \choose r-1} + {n-1 \choose r} & = \frac rn {n\choose r} + \frac {n-r} n {n\choose r} \\ & = {n \choose r} \end {align} $$

as required.

How about this:

Note that

$${n \choose r} = \frac {n(n-1)(n-2)...(n-r+1)} {1\cdot2\cdot3\cdot...(r-1)r}$$

This can be written as $${n \choose r} = \frac nr \cdot \frac {(n-1)(n-2)...(n-r+1)} {1\cdot2\cdot3\cdot...(r-1)} = \frac nr {n-1 \choose r-1} $$ and also $${n \choose r} = \frac n{n-r} \cdot \frac {(n-1)(n-2)...(n-r+1)(n-r)} {1\cdot2\cdot3\cdot...(r-1)r}= \frac n{n-r} {n-1 \choose r}$$

So $$\begin {align} {n-1 \choose r-1} + {n-1 \choose r} & = \frac rn {n\choose r} + \frac {n-r} n {n\choose r} \\ & = {n \choose r} \end {align} $$

as required.