I don't think this question is posed correctly, because it's definitely not true: $30=2\cdot3\cdot5=13+17\Rightarrow s=2$ and $42=2\cdot3\cdot7=5+7+13+17\Rightarrow s=4$ are just two counterexamples.
For the corrected post: We want to show that if $a=p_1p_2p_3=q_1q_2\dots q_s$, then $s=3$. Clearly, $a$ divides $p_1$, so by the theorem on prime division, one of the $q_i$ divides $p_1$. Since $q_i$ is prime, then, $q_i=p_1$ and we can cancel it to get $p_2p_3=r_1r_2\dots r_{s-1}$ with $r_i$ prime. Repeating this process (assuming $s\geq 3$), we get $1=r_1r_2\dots r_{s-3}$, and all of the $r_i$ are integers, so either $s-3=0$ or $|r_i|=1$. But $r_i\geq 2$, so $s=3$. If $s<3$, then we will instead end up as $p_2p_3=1$ or similar. In this case, $p_i>1$ implies a contradiction. Thus $s=3$ is the only possible conclusion.