It's useful to regard these sequences as functions $a,b:\mathbb N\to\mathbb R$ instead, so the derivative of $a$ is given by $a'(i)=a(i+1)-a(i)$. For we will denote the $k-$th derivative of $a$ by $a^{(k)}$, so $a'=a^{(1)}.$ We will also write $a^{(0)}$ for $a$. Note that for any two functions $f,g$, $(fg)'=f'g'+fg'+gf'$; this follows from the identity $$f(i+1)g(i+1)-f(i)g(i)=\left(f(i+1)-f(i)\right)\left(g(i+1)-g(i)\right)+f(i)\left(g(i+1)-g(i)\right)+g(i)\left(f(i+1)-f(i)\right).$$Also, it's easy to see that derivative respects addition and scaling: $(f+g)'=f'+g'$ and $(cf)'=c\cdot f'$. Let $a,b$ be functions corresponding to two good sequences. We claim that for all $k\in\mathbb N$, $(ab)^{(k)}$ is some finite positive linear combination of terms like $a^{(i)}b^{(j)}$ for $i,j$ nonnegative integers. The base case is already done. For the next step, simply note that if $(ab)^{(k)}=\sum c_{i,j}a^{(i)}b^{(j)}$, then its derivative $(ab)^{(k+1)}$ is just $$\sum c_{i,j}\left(a^{(i)}b^{(j)}\right)'=\sum c_{i,j} \left(a^{(i+1)}b^{(j+1)}+a^{(i)}b^{(j+1)}+a^{(i+1)}b^{(j)}\right).$$So we are good, and $(ab)^{(k)}$ is indeed positive for all $k$.