Solution:
First, let $S$ be the sum of all possible products of three distinct variables from the set $\{a, b, c, d, e, f\}$, i.e. $S=abc+abd+abe+\cdots+def$. Now, we assert that
$$\begin{aligned} & (a+b+c+d+e+f)^3 + 2\left(a^3+b^3+c^3+d^3+e^3+f^3\right) \\ & \quad - 3(a+b+c+d+e+f)\left(a^2+b^2+c^2+d^2+e^2+f^2\right) = 6S \end{aligned}$$
To see this, note that, when the left-hand side is expanded, the terms obtained come in three forms: $a^3$ (i.e. cubes of one variable), $a^2b$ (squares of one variable times another), and $abc$ (products of three different variables). Each term of the form $a^3$ has coefficient $1$ in $(a+\cdots+f)^3$, $2$ in $2(a^3+\cdots+f^3)$, and $-3$ in $-3(a+\cdots+f)(a^2+\cdots+f^2)$, for a total coefficient of $0$; the terms of the form $a^2b$ have respective coefficients $3, 0, -3$, for a total of $0$; those of the form $abc$ have coefficients $6, 0, 0$, for a total of $6$. Since the right side of (1) consists of the $abc$-type terms with coefficient $6$, the assertion is proven.

However, the given tells us that the left-hand side of (1) is zero, so $S=0$. Now, let $p=ab-(cd+ce+cf+de+df+ef)$. We claim that the following holds for all numbers $x$:
$$(x+c)(x+d)(x+e)(x+f) = (x-a)(x-b)(x^2-p) + abp + cdef.$$
We can verify this by comparing coefficients for corresponding powers of $x$.

The coefficient of $x^4$ on both sides is $1$; the coefficient of $x^3$ is $c+d+e+f$ on the left and $-(a+b)$ on the right, but the given implies that these are equal. The coefficient of $x^2$ is $cd+ce+cf+de+df+ef$ on the left and $ab-p$ on the right; these are equal by the definition of $p$. The coefficient of $x$ is $cde+cdf+cef+def$ on the left and $ap+bp$ on the right. To see that these are equal, observe that
$$\begin{gathered} cde+cdf+cef+def-(a+b)p = cde+cdf+cef+def-(a+b)ab+(a+b)(cd+ce+cf+de+df+ef) \\ = (S - [abc+abd+abe+abf]) - (a+b)ab = -ab(c+d+e+f+a+b) = 0 \end{gathered}$$
where we have used the definition of $S$ for the second equality and the fact that $S=0$ for the third. Finally, the constant term equals $cdef$ on both sides. Thus, equation (2) holds. Plugging in $x=a$ and $x=b$, we find
$$\begin{aligned} & (a+c)(a+d)(a+e)(a+f) = (a-a)(a-b)(a^2-p) + abp + cdef = abp + cdef \\ & (b+c)(b+d)(b+e)(b+f) = (b-a)(b-b)(b^2-p) + abp + cdef = abp + cdef \end{aligned}$$
Since the right-hand sides match, we are done.