This is a partial answer.
We want to find all the non-zero integer solutions of $\displaystyle\prod_{j=1}^{n}a_j=\sum_{j=1}^{n}a_j$ for $n\ge 2$.
This answer is only interested in the solution $(a_1,a_2,\cdots,a_n)$ satisfying $|a_1|\le |a_2|\le \cdots \le|a_n|$. If $a_j=-a_{j+1}$, then this answer is only interested in $(a_j,a_{j+1})$ satisfying $a_j\lt a_{j+1}$.
(For example, for $n=4$, this answer is interested in $(a_1,a_2,a_3,a_4)=(\color{red}{-1},\color{red}1,-2,-2)$, not $(\color{red}1,\color{red}{-1},-2,-2)$.)
This answer proves the following claims :
Claim 1 : If $n\equiv 1\pmod 4$, then the equation has infinitely many non-zero integer solutions.
Claim 2 : If $n\not\equiv 1\pmod 4$, then the equation has only finitely many non-zero integer solutions.
Claim 3 : If $n\ge 3$, then the equation has at least one non-zero integer solutions where at least one $a_i$ is negative.
Claim 4 : If $n=2$, the only non-zero integer solution of the equation is $(a_1,a_2)=(2,2)$.
Claim 5 : If $n=3$, the only non-zero integer solutions of the equation are $(a_1,a_2,a_3)=(1,2,3),(-1,-2,-3)$.
Claim 6 : If $n=4$, the only non-zero integer solutions of the equation are $(a_1,a_2,a_3,a_4)=(1,1,2,4),(-1,1,-2,-2)$.
Claim 7 : If $n=5$, the only non-zero integer solutions of the equation are $(a_1,a_2,a_3,a_4,a_5)=(-1,-1,1,1,k),(1,1,1,2,5),(-1,-1,-1,-2,-5),(1,1,1,3,3),(-1,-1,-1,-3,-3),(1,1,2,2,2),(-1,-1,-2,-2,-2)$ where $k$ is any non-zero integer.
Claim 8 : If $n=6$, the only non-zero integer solutions of the equation are $(a_1,a_2,a_3,a_4,a_5,a_6)=(1,1,1,1,2,6),(-1,-1,1,1,2,2),(-1,-1,-1,-1,-2,2),(-1,-1,-1,1,-2,-4)$
I'll use the following ideas to prove the claims.
We may suppose that $|a_1|\le |a_2|\le\cdots \le |a_n|$.
$\displaystyle\bigg|\prod_{j=1}^{n-1}a_j\bigg|\le n$.
Proof : $\displaystyle\bigg|\prod_{j=1}^{n-1}a_j\bigg||a_n|=\bigg|\sum_{j=1}^{n}a_j\bigg|\le \sum_{j=1}^{n}|a_j|\le n|a_n|$
For $n\ge 3$, $|a_1|=|a_2|=\cdots =|a_m|=1$ and $\displaystyle\bigg|\prod_{j=m+1}^{n-1}a_j\bigg|\le n$ where $m=n-1-\lfloor\log_2n\rfloor$.
Proof : Suppose that $|a_m|\ge 2$. Then, $n\ge \displaystyle\bigg|\prod_{j=1}^{n-1}a_j\bigg|\ge \bigg|\prod_{j=m}^{n-1}a_j\bigg|\ge 2^{n-m}=2^{1+\lfloor\log_2n\rfloor}\gt 2^{\log_2 n}=n$ which is a contradiction.
$\displaystyle\bigg|\prod_{j=m+1}^{n-1}a_j\bigg|\le n-m+\frac{m}{|a_{n-1}|}$
Proof : $\displaystyle\bigg|\prod_{j=m+1}^{n}a_j\bigg|\le m+(n-m)|a_n|$, so $\displaystyle\bigg|\prod_{j=m+1}^{n-1}a_j\bigg|\le n-m+\frac{m}{|a_n|}\le n-m+\frac{m}{|a_{n-1}|}$
$m+\displaystyle\sum_{j=m+1}^{n}a_j=\prod_{j=m+1}^{n}a_j\pmod 2$
Proof : This is because $a_1\equiv a_2\equiv \cdots\equiv a_m\equiv 1\pmod 2$.
If $a_1a_2\cdots a_{n-1}-1\not=0$, then $\displaystyle 1-\frac{1}{|a_{n-1}|}\sum_{j=1}^{n-1}|a_j|\le\prod_{j=1}^{n-1}a_j\le 1+\frac{1}{|a_{n-1}|}\sum_{j=1}^{n-1}|a_j|$
Proof : $|a_{n-1}|\le |a_n|=\bigg|\dfrac{a_1+a_2+\cdots +a_{n-1}}{a_1a_2\cdots a_{n-1}-1}\bigg|\le\dfrac{|a_1|+|a_2|+\cdots +|a_{n-1}|}{|a_1a_2\cdots a_{n-1}-1|}\implies |a_1a_2\cdots a_{n-1}-1|\le\frac{|a_1|+|a_2|+\cdots +|a_{n-1}|}{|a_{n-1}|}$
If $(a_1,a_2,\cdots, a_n)=(b_1,b_2,\cdots, b_n)$ is a solution where $n$ is odd, then $(a_1,a_2,\cdots, a_n)=(-b_1,-b_2,\cdots, -b_n)$ is a solution.
If $(a_1,a_2,\cdots, a_n)=(b_1,b_2,\cdots, b_n)$ is a solution, then $(a_1,a_2,\cdots, a_{n+4})=(-1,-1,1,1,b_1,b_2,\cdots, b_n)$ is a solution.
In the following, I'm going to prove the claims.
Claim 1 : If $n\equiv 1\pmod 4$, then the equation has infinitely many non-zero integer solutions.
Proof : $(a_1,a_2,\cdots,a_n)=(\underbrace{-1,-1,\cdots,-1}_{\frac{n-1}{2}},\underbrace{1,1,\cdots, 1}_{\frac{n-1}{2}},k)$ is a solution where $k$ is any non-zero integer.
Claim 2 : If $n\not\equiv 1\pmod 4$, then the equation has only finitely many non-zero integer solutions.
Proof : We have $a_n(a_1a_2\cdots a_{n-1}-1)=a_1+a_2+\cdots +a_{n-1}$. Suppose that $a_1a_2\cdots a_{n-1}-1=0$. Then, $|a_1|=|a_2|=\cdots =|a_{n-1}|=1$. Let $A$ be the number of $j(1\le j\le n-1)$ such that $a_j=1$. Let $B$ be the number of $j(1\le j\le n-1)$ such that $a_j=-1$. Since $a_1+a_2+\cdots +a_{n-1}=0$, we see that $B$ is even and that $A=B$. It follows that $n-1=A+B=2B\equiv 0\pmod 4$ which is a contradiction. Since $a_1a_2\cdots a_{n-1}-1\not=0$, we have $|a_{n-1}|\ge 2$ and $|a_n|=\bigg|\frac{a_1+a_2+\cdots +a_{n-1}}{a_1a_2\cdots a_{n-1}-1}\bigg|\le\frac{(n-1)|a_{n-1}|}{|a_1a_2\cdots a_{n-1}-1|}=\frac{n-1}{|a_1a_2\cdots a_{n-2}-\frac{1}{a_{n-1}}|}\le\frac{n-1}{||a_1a_2\cdots a_{n-2}|-\frac{1}{|a_{n-1}|}|}\le \frac{n-1}{\frac 12}=2(n-1)$
Claim 3 : If $n\ge 3$, then the equation has at least one non-zero integer solutions where at least one $a_i$ is negative.
Proof : For odd $n$, $(a_1,a_2,\cdots,a_n)=(-1,-1,\cdots, -1,-2,-n)$ is a solution. If $(a_1,a_2,\cdots, a_n)=(b_1,b_2,\cdots, b_n)$ is a solution where $n$ is even, then $(a_1,a_2,\cdots, a_{n+2})=(-1,1,-b_1,-b_2,\cdots, -b_n)$ is a solution. Since $(a_1,a_2)=(2,2)$ is a solution for $n=2$, the equation for even $n\ge 4$ has at least one non-zero integer solutions where at least one $a_i$ is negative.
Claim 4 : If $n=2$, the only non-zero integer solution of the equation is $(a_1,a_2)=(2,2)$.
Proof : Since $(a_1-1)(a_2-1)=1$ where $a_1,a_2$ are non-zero, we have $(a_1-1,a_2-1)=(1,1)$.
Claim 5 : If $n=3$, the only non-zero integer solutions of the equation are $(a_1,a_2,a_3)=(1,2,3),(-1,-2,-3)$.
Proof : We have $|a_1|=1,|a_2|\le 2+\frac{1}{|a_2|}$ and $|a_2|\le 2$. For $a_1=1$, we have $a_3=1+\frac{2}{a_2-1}$ which implies $a_2=2$ and $a_3=3$.
Claim 6 : If $n=4$, the only non-zero integer solutions of the equation are $(a_1,a_2,a_3,a_4)=(1,1,2,4),(-1,1,-2,-2)$.
Proof : We have $|a_1|=1,|a_2|\le 2, |a_2a_3|\le 3+\frac{1}{|a_3|}$ and $1+a_2+a_3+a_4\equiv a_2a_3a_4\pmod 2$. There is only one odd integer in $a_2,a_3$ and $a_4$. So, we have $|a_2|=1$ and $|a_3|=2$. Since we have $-1\le a_1a_2a_3\le 3$, we get $a_1a_2a_3=2$ and $|a_1+a_2+a_3|\ge 2$.
Claim 7 : If $n=5$, the only non-zero integer solutions of the equation are $(a_1,a_2,a_3,a_4,a_5)=(-1,-1,1,1,k),(1,1,1,2,5),(-1,-1,-1,-2,-5),(1,1,1,3,3),(-1,-1,-1,-3,-3),(1,1,2,2,2),(-1,-1,-2,-2,-2)$ where $k$ is any non-zero integer.
Proof : If $a_1a_2a_3a_4-1=0$, then $(a_1,a_2,a_3,a_4,a_5)=(-1,-1,1,1,k)$ is a solution where $k$ is any non-zero integer. In the following, $a_1a_2a_3a_4-1\not=0$. We have $|a_1|=|a_2|=1,|a_3|\le 2$ and $|a_3a_4|\le 3+\frac{2}{|a_4|}$.
(case 1) If $|a_3|=1$, then $|a_4|\le 3$.
For $|a_4|=1$, $a_5=\pm 1$.
For $|a_4|=2$, since $-\frac 32\le a_1a_2a_3a_4\le \frac 72$, we have $a_1a_2a_3a_4=2$ and $|a_1+a_2+a_3+a_4|\ge 2$.
For $|a_4|=3$, since $-1\le a_1a_2a_3a_4\le 3$, we have $a_1a_2a_3a_4=3$ and $a_1+a_2+a_3+a_4=\pm 6$ since $6=(3-1)\times 3\le |a_1+a_2+a_3+a_4|\le |a_1|+|a_2|+|a_3|+|a_4|=6$
(case 2) If $|a_3|=2$, then $|a_4|=2$. Since $-2\le a_1a_2a_3a_4\le 4$, we have $a_1a_2a_3a_4=4$ and $a_1+a_2+a_3+a_4=\pm 6$.
Claim 8 : If $n=6$, the only non-zero integer solutions of the equation are $(a_1,a_2,a_3,a_4,a_5,a_6)=(1,1,1,1,2,6),(-1,-1,1,1,2,2),(-1,-1,-1,-1,-2,2),(-1,-1,-1,1,-2,-4)$
Proof : We have $|a_1|=|a_2|=|a_3|=1,|a_4|\le 2,|a_4a_5|\le 3+\frac{3}{|a_5|}$ and $1+a_4+a_5+a_6\equiv a_4a_5a_6\pmod 2$. There is only one odd integer in $a_4,a_5$ and $a_6$.
(case 1) If $|a_4|=1$, then $|a_5|=2$. For $a_1a_2a_3a_4a_5=2$, we have $|a_1+a_2+a_3+a_4+a_5|\ge 2$. For $a_1a_2a_3a_4a_5=-2$, we have $a_1+a_2+a_3+a_4+a_5=\pm 6$.
(case 2) If $|a_4|=2$, then $|a_5|=2$. Since $-\frac 52\le a_1a_2a_3a_4a_5\le\frac 92$, we have $a_1a_2a_3a_4a_5=4$ and $|a_1+a_2+a_3+a_4+a_5|\ge 6$.
