Solution:

Yes Bob can. Let the numbers be $v, w, x, y, z$ and let the missing product be $x y z$. The nine given products are $1,2,3,4,5,6,10,12,15$.

Partition these products into the three sequences
$$s_{1}=(v y z, v z x, v x y),\quad s_{2}=(w y z, w z x, w x y),\quad s_{3}=(v w x, v w y, v w z)$$
and call these sequences special. Note that $s'=(y z, z x, x y)$ is directly proportional to the first two and inversely proportional to the last one. The multiples of $5$ thus have to be in the same special sequence, WLOG $(5,10,15)$. This sequence is inversely proportional to at least one of other two special sequences, which would thus have to be a multiple of $(6,3,2)$. However, it cannot be $(6,3,2)$ itself as that would leave $1,4,12$, which cannot form a special sequence. Thus, the multiple of $(6,3,2)$ has to be $(12,6,4)$, leaving $(1,2,3)$ as the final sequence. WLOG, we thus can get the equations
$$((v y z, v z x, v x y),(w y z, w z x, w x y),(v w x, v w y, v w z))=((5,10,15),(1,2,3),(12,6,4)).$$
We can then deduce $x y z = v y z \cdot w x z / v w z = 5 \cdot 2 / 4 = 5 / 2$.

Indeed, $5 / 2$ is achieved by
$$(v, w, x, y, z)=\left(2 \sqrt[3]{15}, \frac{2}{5} \sqrt[3]{15}, \sqrt[3]{15}, \frac{1}{2} \sqrt[3]{15}, \frac{1}{3} \sqrt[3]{15}\right)$$