You can restrict to subsets of the integers to get nice examples of structures that do not have unique factorization.
For example, take the set of all natural numbers, or integers it does not really matter, that are congruent to $1$ modulo $4$. This is a semigroup with identity under multiplication, and every element is the product of irreducible elements. However, factorizations into irreducible elements are not unique in an essential way. For example, you have:
$$21 \cdot 21 = 9 \cdot 49$$
and $9,21,49$ are irreducible in that structure as their divisors are $3$ modulo $4$.
This structure is sometimes called Hilbert semigroup as he used (it is said) this as instructional example. (Added: this is also mentioned in another answer that appeared while I was writing this.)
Another possibility would be to consider only the even integers (and $1$). Again you have a factorization into irreducible elements, but it is not unique;
$$42 \cdot 2 = 6 \cdot 14$$
Still another one would be to consider all integers greater than $2$ (and $0$) in this case with addition as operation. The only two irreducible elements are $2$ and $3$ and one has non-unique factorizations:
$$2+2+2 = 3 + 3$$
The above is an example of a numerical semigroup.
If you do not like this is additive, you can instead consider polynomials that do not have a linear term.
I am not sure if this is relevant in your context, but if you want you could continue to use some of these examples in more advanced contexts. For instance, the Hilbert semigroup is contained in the (odd) natural numbers and that structure has unique factorization. This situation is analogous, in fact not just analogous but in a suitable sense littelraly the same, as one encounter when studying factorizations into ideals to restore unique factorization in algebraic number theory.