proof verification - Elementary demonstration; $p$ prime, $1 lt a lt p$, $;1 lt b lt p quad$ Then $ pnmid a b$

Update: If this question is of interest, you can also click here.

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Update: Using Bill Dubuque's lemma and logic proving Euclid's lemma, we can supply an elementary proof.

To get a contradiction, assume than $p \mid a b$.

Let $S = \{n \in \Bbb N \, | \, p \mid nb \}$. Then $p \in S$ and $a \in S$. Moreover, $S$ is closed under subtraction.

Let $d = \text{min(}S\text{)}$. By the lemma, $d \mid p$, so $d = 1$ or $d = p$.

If $d = 1$, since $d \in S$, it must follow that $p \mid (1 \times b)$, which is absurd since $b \lt p$.

By the lemma, $d \mid a$, so if $d = p$ then $p \mid a$, which is absurd since $a \lt p$.

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I've been motivated (see this) to prove the following result using only elementary techniques.

Let $p$ be a prime greater than $2$.

Let $1 \lt a \lt p$

Let $1 \lt b \lt p$

Then

$$\tag 1 p\nmid a b$$

I think that this is as simple as first showing that

$$ \text{For every integer n } \ge 1 \text{ such that } p\nmid n, \; \; p\nmid na$$

and ironing out some details.

Using only the 'first page' of elementary theory of the natural numbers/integers (for example Euclidean division, the construction of $\Bbb Z$, the existence of prime factorizations and that modular arithmetic is well-defined), can this approach work for proving $\text{(1)}$?

Besides answering yes in the comments, a proof would be appreciated (this elementary approach can be exhausting).