UPDATE/WARNING: DO NOT READ (WASTE OF TIME)
The effort I put in here is now an embarrassment as it goes nowhere. I can't delete this posting since there is an answer, but if a moderator could delete it I would be much obliged.
I deleted all my work here (including comments and answers) and have rolled the question back to the original posting, except for this update.
Let $n$ be any integer greater than $1$.
Let $p$ be any prime factor of $n$.
Consider any factorization of $n$,
$\tag 1 n = ab$
Write both $a$ and $b$ using their $\text{Base-p}$ representations.
$\tag 2 a = a_0 + a_1 p + a_2 p^2 + a_3 p^3 + \dots$
$\tag 3 b = b_0 + b_1 p + b_2 p^2 + b_3 p^3 + \dots$
By the elementary theory found here, the only way $p$ can divide the product $ab$
$\quad$ = rhs of $\text{(2)}$ $\times$ rhs of $\text{(3)}$
is if $a_0 = 0$ or $b_0 = 0$.
Summarizing,
Proposition 1: Let $n$ be any integer with a prime factor $p$ and let $n = ab$.
Then $p$ must divide either $a$ or $b$.
Continuing, we know that if $n \gt 1$ the first integer greater than $1$ that divides into $n$, $\gamma(n)$, is a prime, the minimal prime factor.
Any integer $n$ can be factored in prime numbers, but an easy argument using proposition 1 guarantees that $\gamma(n)$ must appear in any factorization.
Theorem 2: Any two factorizations of an integer $n \gt 1$ into primes must agree.
Proof
Let
$\tag 4 n = \gamma(n)\, p_1 p_2 \dots$
$\tag 5 n = \gamma(n)\, q_1 q_2 \dots$
be any two factorizations. We can factor out $\gamma(n)$ in both representations. Now each remaining representation is either empty or contains more factors. If necessary, then, repeat, factoring out
$\gamma(\frac{n}{\gamma(n)})$
In this way all factors are matched up and uniqueness is guaranteed. $\quad \blacksquare$
Well, I didn't bother reading the linked stuff, but there are certainly omissions here.
For instance, talking about "the first integer greater than $1$ that divides into $n$" should probably be supported with two facts: the set $S$ of integers greater-than-one dividing $n$ contains $n$, hence is nonempty, hence, by the well-ordering principle, has a least element.
Then you also need to make the point that this least element is indeed prime.
The claim that "any integer can be factored into primes" again requires some proof (and again, the well-ordering principle might want to be involved). You probably want to talk about positive integers here, too9.
For your claim that any two factorizations "agree," I think you need to define "agreement." For instance, 2 is the smallest prime factor of 12, but
don't "agree", at least not term-by-term. So there's more work to do here.
If you're going to give an elementary proof, you need to actually prove everything.
One example of something that needs proving: that a (positive) integer has a unique base-$p$ representation. That claim might well depend on FTA (although I doubt it, offhand). Anyhow, without a citation to a proof, I have to regard your very starting point as suspect.
In short: you've got an OK idea here, but not something I'd accept as an elementary proof.