Ok, I have a functional generalization of your Product-Sum conjecture using a very simple method.
Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be any function with a nonnegative $\binom{n}{2}$th derivative. I claim that we have the following functional inequality:
$\sum_{\pi\in S_n} (-1)^{\sigma(\pi)}f(\sum_i a_ib_{\pi(i)}) \ge 0$.
Plugging in $f(x) = -(-1)^{\binom{n}{2}}\log(x)$, we see that your inequality holds as long as $\sum_i a_ib_{n+1-i} \ge 0$.
To prove the functional inequality, it clearly suffices to prove it in the case that the $b_i$s are all distinct positive integers, so assume from now on that this is the case. Let $x_i = e^{a_i}$. Note first that in the special case in which $f(x) = e^x$, we get that
$\sum_{\pi\in S_n} (-1)^{\sigma(\pi)}\prod_ix_i^{b_{\pi(i)}} = \det((x_i^{b_j})_{i,j})$
which is $\prod_{i\lt j}(x_i-x_j)$ times the Schur polynomial $s_{(b_1-n+1,...,b_n)}(x_1,...,x_n)$, and as is well known Schur polynomials have all of their coefficients nonnegative. For every monomial $x^m = \prod_i x_i^{m_i}$, let $c_m$ be its coefficient in the Schur polynomial $s_{(b_1-n+1,...,b_n)}(x_1,...,x_n)$.
Next, let $S_a$ be the shift operator - i.e., let $S_a(f)(x) = f(x+a)$. Then it is easy to check that we have
$\sum_{\pi\in S_n} (-1)^{\sigma(\pi)}f(\sum_i a_ib_{\pi(i)}) =\sum_mc_m(\prod_{i\lt j}(S_{a_i}-S_{a_j}))(f)(\sum_ia_im_i)$,
which is nonnegative since we have $(\prod_{i\lt j}(S_{a_i}-S_{a_j}))(f)(x) \ge 0$ for any $x$ and any function $f$ with nonnegative $\binom{n}{2}$th derivative.
Best Answer
Without the condition that all coefficients are positive there is the following ounterexample: $$a(x)=(x+1)(x+2)(x+50).$$ For $a(x)-tx$ all zeros are real when $t\leq 0$ and when $t>300$. But for $t\in (100,200)$ only one zero is real. Two zeros disappeared from the negative ray, and after some time re-appear on the positive ray. Take $b(x)=a(x)-400x.$
Here is the graph of $a(x)/x$:
http://www.math.purdue.edu/~eremenko/dvi/graph.pdf