For faster typing, I'm going to label the roots of the first equation $\{m, n, p\}$, and of the second equation $\{t,u,v\}$. So we have:
$$
\begin{gather}
m+n+p = tuv = -a \\
mnp = t+u+v = -c \\
mn+mp+np=tu+tv+uv = 0
\end{gather}
$$
A lot of the following requires assuming WLOG, as $\{m, n, p\}$ and $\{t, u, v\}$ are all interchangeable within their own sets, and the sets themselves are swappable.
We can see that $a, c \neq 0$. If $a=0$, then at least one of $t,u,v = 0$. Choose $t=0$. Then
$$tu+tv+uv = 0u+0v+uv = 0 \implies uv=0$$
That means $t = u = 0$, which doesn't give us three distinct roots. The same applies to $c$. Therefore, each of $a,c$ must have three different (not necessarily prime) factors.
To have a solution, we need two distinct sets of three distinct integers. Each of them must have a sum equal to the other's product, and the sum of the pairwise products must be equal to zero. That last bit means each set must have at least one negative and at least one positive integer. For the sake of simplicity, we will choose $m>0, p<0$ and $t>0, v < 0$.
Therefore, we need two sets such that $mnp = t+u+v$ and $tuv = m+n+p$. To have that, we require:
$$|mnp| > |m+n+p| \iff |tuv| < |t+u+v|$$
(The values cannot be equal, or our two sets will be the same.) And here's our wall: there are no three distinct integers $t, u, v$ such that $|t+u+v| > |tuv|$. There are cases where they are equal, e.g., $1+2+3 = 1 \cdot 2 \cdot 3$ and $|(-1)(1)(k)| = |k+1-1|$, but none where the sum is greater.
So, alas, your intuition was incorrect. As was mine, as I thought I'd find at least one solution. Hope this helps!
(Brief side note: if we don't require distinct integers, $\{1, 1, k\}$ and $\{1,2,2\}$ have sums less than their product. I believe those are the only sets that work, though.)
Best Answer
Let us first examine some examples to guess the general case:$$(x-r_1)(x-r_2)=x^2-(r_1+r_2)x+r_1r_2$$ $$(x-r_1)(x-r_2)(x-r_3)=x^3-(r_1+r_2+r_3)x^2+(r_1r_2 + r_1r_3+r_2r_3)x-r_1r_2r_3.$$So, we can guess the following identity:$$\prod_{i=1}^n(x-r_i)=\sum_{k=0}^n\sum_{1 \le j_1 \lt ... \lt j_k \le n}(-1)^kr_{j_1} ... r_{j_k}x^{n-k}.$$Let us prove the claim by induction.
The base case is trivial. So, let us assume that the claim is correct for $n=m$, that is,$$\prod_{i=1}^m(x-r_i)=\sum_{k=0}^m\sum_{1 \le j_1 \lt ... \lt j_k \le m}(-1)^kr_{j_1} ... r_{j_k}x^{m-k}.$$ So, we need to prove the claim for $n=m+1$ as follows.$$\prod_{i=1}^{m+1}(x-r_i)=\left ( \prod_{i=1}^m(x-r_i) \right ) \left ( \vphantom{\prod_{i=}^n} x-r_{m+1} \right )$$ $$=\left (\sum_{k=0}^m\sum_{1 \le j_1 \lt ... \lt j_k \le m}(-1)^kr_{j_1} ... r_{j_k}x^{m-k} \right ) \left ( \vphantom{\prod_{i=}^n} x-r_{m+1} \right )$$ $${=\sum_{k=0}^m\sum_{1 \le j_1 \lt ... \lt j_k \le m}(-1)^kr_{j_1} ... r_{j_k}x^{(m+1)-k} -\sum_{k=0}^m\sum_{1 \le j_1 \lt ... \lt j_k \le m}(-1)^kr_{j_1} ... r_{j_k}r_{m+1}x^{m-k}}$$ $$=\left (x^{m+1}+\sum_{k=1}^m\sum_{1 \le j_1 \lt ... \lt j_k \le m}(-1)^kr_{j_1} ... r_{j_k}x^{(m+1)-k} \right ) - \left ( \sum_{k=0}^{m-1}\sum_{1 \le j_1 \lt ... \lt j_k \le m}(-1)^kr_{j_1} ... r_{j_k}r_{m+1}x^{m-k}+ (-1)^m r_{j_1} ... r_{j_m}r_{m+1} \right )$$ $$=\left (x^{m+1}+\sum_{k=1}^m\sum_{1 \le j_1 \lt ... \lt j_k \le m}(-1)^kr_{j_1} ... r_{j_k}x^{(m+1)-k} \right ) - \left ( \sum_{k=1}^{m}\sum_{1 \le j_1 \lt ... \lt j_k \le m}(-1)^{k-1}r_{j_1} ... r_{j_{k-1}}r_{m+1}x^{m-(k-1)}+ (-1)^m r_{j_1} ... r_{j_m}r_{m+1} \right )\tag{*}\label{*}$$ $$= \left ( x^{m+1} + (-1)^m r_{j_1} ... r_{j_m}r_{m+1} + \sum_{k=1}^m (-1)^k \left ( \sum_{1 \le j_1 \lt ... \lt j_k \le m}r_{j_1} ... r_{j_k}+ r_{j_1} ... r_{j_{k-1}}r_{m+1} \right ) x^{(m+1)-k} \right )$$ $${= \left ( x^{m+1} + (-1)^m r_{j_1} ... r_{j_m}r_{m+1} + \sum_{k=1}^m (-1)^k \sum_{1 \le j_1 \lt ... \lt j_k \le m+1}r_{j_1} ... r_{j_k} x^{(m+1)-k} \right )}$$ $$=\sum_{k=0}^{m+1} \sum_{1 \le j_1 \lt ... \lt j_k \le m+1}(-1)^k r_{j_1} ... r_{j_k} x^{(m+1)-k}.\tag{**}\label{**}$$Thus, by induction we proved that for any natural number $n$ the following identity holds:$$\prod_{i=1}^n(x-r_i)=\sum_{k=0}^n\sum_{1 \le j_1 \lt ... \lt j_k \le n}(-1)^kr_{j_1} ... r_{j_k}x^{n-k}.$$
Footnote
\ref{*} is followed from the following property of summation:$$\sum_{i=m}^nA_i=\sum_{i=m+1}^{n+1}A_{i-1}.$$ \ref{**} is followed from considering the fact that for any fixed $k$ one can decompose the sum $\sum_{1 \le j_1 \lt ... \lt j_k \le m+1} r_{j_1} ... r_{j_k}$ into two sums: (i) the sum of terms not containing $r_{j_{m+1}}$, that is, $\sum_{1 \le j_1 \lt ... \lt j_k \le m} r_{j_1} ... r_{j_k}$, and (ii) the sum of terms containing $r_{j_{m+1}}$, that is $\sum_{1 \le j_1 \lt ... \lt j_k \le m} r_{j_1} ... r_{j_{k-1}}r_{j_{m+1}}$.