Let $b_{i}=a_{1} a_{2} \cdots a_{i}$, for $1 \leq i \leq p$. We prove that it is possible to choose the permutation such that $b_{i} \equiv i \bmod p$ for all $i$. For $i \geq 2$, $a_{i} \equiv b_{i} \cdot b_{i-1}^{-1} \bmod p$ if $b_{i-1} \not \equiv 0 \bmod p$. We now choose $a_{1}=1$ and $a_{i} \equiv i \cdot(i-1)^{-1} \bmod p$ for $2 \leq i \leq p$. It is now sufficient to prove that $a_{i} \not \equiv 1 \bmod p$ for all $2 \leq i \leq p$ and $a_{i} \not \equiv a_{j} \bmod p$ for all $2 \leq j<i \leq p$.
Assume for the sake of contradiction that $a_{i} \equiv 1 \bmod p$ for some $2 \leq i \leq p$. Then $i \cdot(i-1)^{-1} \equiv 1$ $\bmod p$, so $i \equiv i-1 \bmod p$, so $0 \equiv -1 \bmod p$. Since $p \geq 2$, this is a contradiction. Now assume that $a_{i} \equiv a_{j} \bmod p$ for some $2 \leq j<i \leq p$. Then $i \cdot(i-1)^{-1} \equiv j \cdot(j-1)^{-1}$ $\bmod p$ so $i(j-1) \equiv j(i-1) \bmod p$ so $ij - i \equiv ij - j \bmod p$ so $-i \equiv -j \bmod p$. But we had $2 \leq j < i \leq p$, so this cannot be.
We conclude that if we choose the $a_{i}$ as indicated above, all $a_{i}$ will be distinct, making it indeed a permutation of $1,2, \ldots, p$. Furthermore, by definition, $a_{1} a_{2} \cdots a_{i} \equiv i \bmod p$, so the second condition is also satisfied.