Solution:

First, note that the product and the quotient (when this is an integer) of two perfect cubes is again a perfect cube. The quantity
$$\frac{(ab)(cd)(ef)(ga)}{(bc)(de)(fg)} = a^{2}$$
is then a perfect cube. Now, if $a^{2}$ is a perfect cube, then $a$ is also a perfect cube: indeed, $a^{2}$ being a perfect cube means that, letting $a = p_{1}^{t_{1}} \cdots p_{r}^{t_{r}}$ be the prime factorization of $a$, $2 t_{i}$ is a multiple of 3 for every $i$; but then $t_{i}$ is also a multiple of 3 for every $i$, which means that $a$ is a perfect cube. We thus know that $a$ and $ab$ are perfect cubes; it follows that $b$ is also one. We now know that $b$ and $bc$ are perfect cubes, from which it follows that $c$ is also one, and so on.

Note that, for a number, being a perfect cube means that for every prime $p_{i}$ appearing with exponent $t_{i}$ in its factorization, $t_{i}$ is a multiple of 3. Let us then fix a prime $p$ and call $t_{a}, t_{b}, \ldots, t_{g}$ the exponents with which it appears in the factorization of $a, b, \ldots, g$. The conditions of the problem tell us that
$$t_{a} + t_{b},\ t_{b} + t_{c},\ t_{c} + t_{d},\ t_{d} + t_{e},\ t_{e} + t_{f},\ t_{f} + t_{g},\ t_{g} + t_{a}$$
are multiples of 3.
Suppose for contradiction that one of the $t$'s is not a multiple of 3. Up to cyclically renaming the integers involved, we may assume it is $t_{a}$. Suppose that the remainder of the division of $t_{a}$ by 3 is 1 (the case in which it is 2 is handled in exactly the same way). Then, from the fact that the first of the expressions above is a multiple of 3, it follows that the remainder of the division by 3 of $t_{b}$ is 2; hence, from the second it follows that $t_{c}$ has remainder 1, and then similarly that $t_{d}$ has remainder 2, $t_{e}$ has remainder 1, $t_{f}$ has remainder 2, and $t_{g}$ has remainder 1. But the last relation says that $t_{g} + t_{a}$ is a multiple of 3, whereas we have shown that both $t_{a}$ and $t_{g}$ have remainder 1, which is impossible. This completes the proof by contradiction.

The fact that the quantities in (1) are multiples of 3 can be expressed as the following linear system over the field of integers modulo 3.
$$\begin{aligned} t_{a} + t_{b} &\equiv 0 \pmod{3} \\ t_{b} + t_{c} &\equiv 0 \pmod{3} \\ t_{c} + t_{d} &\equiv 0 \pmod{3} \\ t_{d} + t_{e} &\equiv 0 \pmod{3} \\ t_{e} + t_{f} &\equiv 0 \pmod{3} \\ t_{f} + t_{g} &\equiv 0 \pmod{3} \\ t_{g} + t_{a} &\equiv 0 \pmod{3} \end{aligned}$$
This system has as its associated matrix the matrix
$$\left(\begin{array}{lllllll} 1 & 1 & & & & & \\ & 1 & 1 & & & & \\ & & 1 & 1 & & & \\ & & & 1 & 1 & & \\ & & & & 1 & 1 & \\ & & & & & 1 & 1 \\ 1 & & & & & & 1 \end{array}\right).$$
Over the field of integers modulo 3, this matrix is invertible (because it has nonzero determinant), so the system has only the trivial solution $t_{a} = t_{b} = \ldots = t_{g} = 0$.