A reduced residue system mod $m$ consists of those congruence classes that are relatively prime to $m$. Stated differently, we say that $r_1, \ldots, r_k$ is a reduced residue system mod $m$ if it is a reordering of every number (mod $m$) from $1$ to $m-1$ that is relatively prime to $m$.
So the two lists of numbers $r_1, r_2, \ldots, r_k$ and $ar_1, ar_2, \ldots, ar_k$ are the same lists of numbers (mod $m$, that is). Therefore multiplying them together mod $m$ must give the same product, which is the step you are asking about.
Let's do an example. Suppose we're interested in showing $2^6 \equiv 1 \pmod 9$. Then a reduced residue system mod $9$ is $$1,2,4,5,7,8.$$
What happens when we multiply each of these by $2$? At first, we get
and after reducing mod $9$ we get
Notice that this is exactly the same as the first list, but in a different order.
Since these are the same lists, their products must be equal. Here, that means that
$$ 1\cdot 2 \cdot 4 \cdot 5 \cdot 7 \cdot 8 \equiv 2 \cdot 4 \cdot 8 \cdot 10 \cdot 14 \cdot 16 \equiv 2^6 \cdot 1 \cdot 2 \cdot 4 \cdot 5 \cdot 7 \cdot 8 \pmod 9.$$
After multiplying by the modular inverse of $1 \cdot 2 \cdot 4 \cdot 5 \cdot 7 \cdot 8$, we see that
$$ 2^6 \equiv 1 \pmod 9.$$
Do you see how it works now?