The unique factorisation theorem for positive integers states that every positive integer can be uniquely expressed as a product of primes.
What does "uniqueness" mean here? Let $n\in \mathbb{Z}^{+}$ and $n = {p_1} {p_2} \ldots {p_r}= {q_1}{q_2}\ldots {q_s}$ be any two factorisations of $n$ as product of primes. Then the uniqueness here simply means that the multisets $\left[p_1, p_2, \ldots, p_r \right]= \left[q_1,q_2, \ldots , q_s\right]$
So far so good. Now, I was wondering if there's a way to write the unique factorisation for any non-zero integer $n$. I guess you may write the prime factorisation of $n<0$ as the prime factorisation of $|n|$ with a "minus sign" but I'm not sure what exactly would remain unique about it then. Because $-1 = (-1)^3=(-1)^5 = \ldots$. (The last part is similar to prime factorisation of $n≠0$ discussed in Chapter 1, Theorem 1 of A Classical Introduction to Modern Number Theory by Ireland and Rosen).
For the record, unlike Ireland & Rosen where it takes that: If $p$ is a prime then $-p$ is also prime, I define primes to be positive integers having two distinct positive divisors, $1$ and $p$ to avoid running into problems like $6= 2 \cdot 3 =(-2)(-3)$ for example. But is there a more natural generalisation of unique factorisation theorem to every non-zero integer? I guess I should also mention that I have not studied Ring Theory yet, just in case. So layman language would be very appreciated.
If you go on to study "abstract algebra" you will encounter the natural generalization you are looking for. Here's a brief look ahead.
A ring is a mathematical structure where addition and multiplication make sense and follow the usual rules. So the set $\mathbb{Z}$ of integers is ring, and so is the set $\mathbb{C}[z]$ of polynomials with complex coefficients.
A unit in a ring is an element whose reciprocal is also in the ring. The units in $\mathbb{Z}$ are $\pm 1$; the units in $\mathbb{C}[z]$ are the nonzero constant polynomials.
In an integral domain a nonunit $p$ is irreducible if whenever $p=rs$, one of $r$ or $s$ must be a unit. It's prime if when $p$ divides $ab$ it divides $a$ or $b$.
In the integers and a polynomial ring over a field these are equivalent, which is one of the key properties that makes unique factorization work:
The fundamental theorem of arithmetic says that every nonzero nonunit in the integers is uniquely a product of primes, where uniqueness means "up to order and to multiplication of the factors by units".
The fundamental theorem of algebra says that every complex polynomial of degree $n > 0$ has $n$ roots when you count them with the right multiplicities. Since roots correspond to linear factors, that says every nonconstant polynomial is "uniquely a product of primes" in the same sense.
You will also encounter interesting rings where the "fundamental theorem" fails because irreducible elements might not be prime.