Here is a tentative proof:
Let $a$ be the number of odd elements in the set $\{a_i; 1\le i \le n \}$ and $b$ the number of odd elements in the set $\{b_i; 1\le i \le n \}$, then, since $\sum_i a_i =\sum_i b_i$,we have $ a \equiv b \bmod 2$. The number of odd summands in the sums corresponding to $k=2$ is $a\choose2 $ and $b \choose 2$ respectively, and then ${a \choose 2} \equiv {b \choose 2} \bmod 2$ and then $a(a-1) \equiv b(b-1) \bmod 4$ and then $a \equiv b \bmod 4$.
Nota: generally, for $ 1 \le k \le n-1$, we have${a \choose k} \equiv {b \choose k} \bmod 2$.
If $n=3$, since $0 \le a,b \le 3$, we then have $a=b$.
When $a=b=0$ or $a=b=1$, clearly $a_1a_2a_3 \equiv 0 \bmod 4$ and $b_1b_2b_3 \equiv 0 \bmod 4$ and $a_1a_2a_3 -b_1b_2b_3 \equiv 0 \bmod 4$.
When $a=b=2$, suppose that $a_1, a_2$ and $b_1, b_2$ are odd and $\alpha_i=\frac{a_i-1}{2}$, resp. $\beta_i=\frac{b_i-1}{2}$.
We have $a_3(a_1+a_2)+ a_1a_2=b_3(b_1+b_2)+ b_1b_2$, then $a_1a_2 \equiv b_1b_2 \bmod 4$, then $(2\alpha_1+1)(2\alpha_2+1)\equiv (2\beta_1+1)(2\beta_2+1) \bmod 4$
then $2\alpha_1+2\alpha_2 \equiv 2\beta_1 +2\beta_2 \bmod 4$
but on the other hand:
$a_3^3+ (2\alpha_1+1)^3+ (2\alpha_2+1)^3 \equiv 6\alpha_1+6\alpha_2+2 \bmod 4 $
$b_3^3+ (2\beta_1+1)^3+ (2\beta_2+1)^3 \equiv 6\beta_1+6\beta_2+2 \bmod 4 $
then $a_3^3+ a_1^3+ a_2^3 -b_3^3-b_1^3-b_2^3 \equiv 0 \bmod 4 $,
then $a_1a_2a_3 -b_1b_2b_3 \equiv 0 \bmod 4$ (from the OP remark).
When $a=b=3$, then $a_3^3+ a_1^3+ a_2^3 -b_3^3-b_1^3-b_2^3 \equiv 6(\alpha_3+ \alpha_1+ \alpha_2 -\beta_3-\beta_1-\beta_2) \bmod 4 $,
but $2\alpha_1 +1 +2\alpha_2 +1+2\alpha_3 +1 = 2\beta_1 +1 +2\beta_2 +1+2\beta_3 +1$ , then $\alpha_1 +\alpha_2 +\alpha_3 =\beta_1 +\beta_2 +\beta_3 $ and then, again by the OP remark, we have $a_1a_2a_3 -b_1b_2b_3 \equiv 0 \bmod 4$.
It is possible to derive a similar proof for $n \gt3$. We first show that $a=b$, in the general case:
We have $a=b$ if $1 \le a\le b \le n-1 $, otherwise if $a \lt b $, we would have $0= {a \choose b} \equiv {b \choose b} =1$.
And if $b=n$, all $b_i$ are odd and we have ${n\choose k} \equiv {a \choose k} \bmod 2$ for all $k$ such that $0 \le k \le n-1$. Suppose that $1 \le a \le n-1$ then in the ring of polynomials of $\mathbb{Z}/2\mathbb{Z}$ we have $\sum_{k=0}^{n-1}{n\choose k}X^k = \sum_{k=0}^{n-1}{a\choose k}X^k$, that is $(1+X)^n -X^n =(1+X)^a$ and $1$ would be a root on the rhs but not on the lhs, which is impossible, then $a=n=b$.
There remains to show that the congruence $\bmod 4$ holds in the two non-trivial cases $a=b=n$ and $a=b=n-1$.
From the Newton-Girard formulae relating sums of powers and elementary symetric polynomials we see that
$\sum_i a_i^k =\sum_i b_i^k$, for $1\le k \le n-1$
and
$\sum_i a_i^n +n(-1)^na_1\cdot\cdot a_n =\sum_i b_i^n +n(-1)^nb_1\cdot\cdot b_n $.
When $a=b=n$, and $n$ is odd, we have $\sum_i a_i^n-\sum_i b_i^n=\sum_i (2\alpha_i+1)^n-\sum_i (2\beta_i+1)^n \equiv 2 (\sum_i \alpha_i -\sum_i \beta_i) =0 \bmod 4$.
Then $a_1\cdot\cdot a_n \equiv b_1\cdot\cdot b_n \bmod 4$.
When $a=b=n$, and $n$ is even, we have $$\frac{\sum_i a_i^n-\sum_i b_i^n}{n}=\frac{\sum_i (2\alpha_i+1)^n}{n}-\frac{\sum_i (2\beta_i+1)^n}{n} =\sum_{1\le i,j \le n}{n-1 \choose j-1} \frac{2^j}{j}(\alpha_i^j-\beta_i^j),$$
that is
$$\frac{\sum_i a_i^n-\sum_i b_i^n}{n}=\sum_{i=1}^n2(\alpha_i-\beta_i)+2(n-1)\sum_{ i =1}^ n(\alpha_i^2-\beta_i^2)+\sum_{ j=3}^ n\sum_{i=1} ^n {n-1 \choose j-1} \frac{2^j}{j}(\alpha_i^j-\beta_i^j),$$
hence $$\frac{\sum_i a_i^n-\sum_i b_i^n}{n} \equiv 2(n-1)\sum_{ i =1}^ n(\alpha_i^2-\beta_i^2) \bmod 4,$$
since $\frac{2^j}{j} \equiv 0 \bmod 4$ for $j\ge 3$.
But $\sum_{ i =1}^ n(\alpha_i^2-\beta_i^2)=0 $ and then $\frac{\sum_i a_i^n-\sum_i b_i^n}{n} \equiv 0 \bmod 4$ and finally $a_1\cdot\cdot a_n \equiv b_1\cdot\cdot b_n \bmod 4$.
And when $a=b=n-1$ and $n$ is odd, we suppose that $a_n$ and $b_n$ are the only even numbers and we have,
$\sum_{i\le n} a_i^n-\sum_{i\le n} b_i^n \equiv \sum_{i\le n-1} (2\alpha_i+1)^n-\sum_{i\le n-1} (2\beta_i+1)^n \equiv n\sum_{i\le n-1}(a_i -b_i) \bmod 4$
but $$a_n\sum_{j=1} ^{n-1}\prod_{1\le i\le n-1, i\neq j}a_i+ a_1\cdot \cdot a_{n-1}=b_n\sum_{j=1} ^{n-1}\prod_{1\le i\le n-1, i\neq j}b_i+ b_1\cdot \cdot b_{n-1}$$
then $$a_1\cdot \cdot a_{n-1} \equiv b_1\cdot \cdot b_{n-1} \bmod 4$$
then, similarly to the case $n=3$ above, we have $ 2\sum_{j=1} ^{n-1}\alpha_i \equiv 2\sum_{j=1} ^{n-1}\beta_i \bmod 4 $ and eventually $ \sum_{j=1} ^{n-1}a_i \equiv \sum_{j=1} ^{n-1}b_i \bmod 4 $ and $\sum_{j=1} ^{n}a_i^n \equiv \sum_{j=1} ^{n}b_i^n \bmod 4 $ and $a_1\cdot \cdot a_{n} \equiv b_1\cdot \cdot b_{n} \bmod 4$, since $n$ is odd.
And when $a=b=n-1$ and $n$ is even, we still suppose that $a_n$ and $b_n$ are the only even numbers and we have,
$$\frac{\sum_i a_i^n-\sum_i b_i^n}{n}=\sum_{i=1}^{n-1}2(\alpha_i-\beta_i)+2(n-1)\sum_{ i =1}^ {n-1}(\alpha_i^2-\beta_i^2)+\sum_{ j=3}^ n\sum_{i=1} ^{n-1} {n-1 \choose j-1} \frac{2^j}{j}(\alpha_i^j-\beta_i^j) + \frac{a_n^n-b_n^n}{n},$$
and reducing modulo $4$ we obtain
$$\frac{\sum_i a_i^n-\sum_i b_i^n}{n} \equiv \sum_{i=1}^{n-1}2(\alpha_i-\beta_i)+2(n-1)\sum_{ i =1}^ {n-1}(\alpha_i^2-\beta_i^2) \bmod 4.$$
But again similarly to the odd $n$ case above, we have $ 2\sum_{j=1} ^{n-1}\alpha_i \equiv 2\sum_{j=1} ^{n-1}\beta_i \bmod 4 $ and then
$$\frac{\sum_i a_i^n-\sum_i b_i^n}{n} \equiv 2(n-1)\sum_{ i =1}^ {n-1}(\alpha_i^2-\beta_i^2) \bmod 4,$$
$$\frac{\sum_i a_i^n-\sum_i b_i^n}{n} \equiv \frac{n-1}{2}\sum_{ i =1}^ {n-1}((a_i-1)^2-(b_i-1)^2) \bmod 4,$$
$$\frac{\sum_i a_i^n-\sum_i b_i^n}{n} \equiv -\frac{n-1}{2}(a_n^2-b_n^2) =-(n-1)\frac{a_n+b_n}{2}(a_n-b_n) \equiv 0 \bmod 4,$$
since $a_n$ and $b_n$ are even and $a_n-b_n =-\sum_{i \le n-1} (a_i-b_i) \equiv 0 \bmod 4$,
and finally, in this last case, we also have $$a_1\cdot \cdot a_{n} \equiv b_1\cdot \cdot b_{n} \bmod 4.$$
