This answer assumes that we may use the followings :
All rational numbers are either terminating decimal or repeating decimal numerals. (see here)
A number has a terminating decimal expansion if and only if it is rational and when in lowest terms, its denominator is coprime to all primes other than $2$ and $5$. (see here)
We want to prove $(1)\iff (2)$ where
$(1)$ A reduced fraction's denominator has factors of $2$ or $5$ or both besides other prime factors.
$(2)$ The decimal representation has some non-repeating digits after the decimal point.
We prove $(1)\iff (2)$ using the following lemma :
(A proof of the lemma is written at the end of this answer)
Lemma : A reduced fraction's denominator is coprime to $10$ if and only if the decimal representation is periodic and the repeating digits appear right after the decimal point.
Proof that $(1)\implies (2)$ :
Since the denominator is not of the form $2^a5^b$, the decimal representation is periodic. Since the denominator is not coprime to $10$, it follows from the lemma that the repeating digits do not appear right after the decimal point. So, the decimal representation has some non-repeating digits after the decimal point.$\ \square$
Proof that $(2)\implies (1)$ :
It follows from the lemma that the reduced fraction's denominator is not coprime to $10$. Suppose that the denominator is of the form $2^a5^b$. Then, it has a terminating decimal expansion, which contradicts that the decimal representation is periodic. So, the denominator has factors of $2$ or $5$ or both besides other prime factors.$\ \square$
Finally, let us prove the lemma :
Lemma : A reduced fraction's denominator is coprime to $10$ if and only if the decimal representation is periodic and the repeating digits appear right after the decimal point.
Proof of the lemma :
("if" part)
Let $r'=\overline{r_1r_2\cdots r_t}$ be the repeating digits and $N$ be the integer part. Then, $f$ can be written as $f=N+\dfrac{r'}{10^t-1}=\dfrac{N(10^t-1)+r'}{10^t-1}$ whose denominator is coprime to $10$.
("only if" part)
Since the reduced fraction's denominator is coprime to $10$, the decimal representation is periodic.
We may suppose that $f=\dfrac nd\lt 1$ and that $f=\dfrac nd=\overline{0.b_1b_2\cdots b_s[R][R]\cdots}$ where $[R]=\overline{r_1r_2\cdots r_t}$ represents repeating digits.
Let $r_0$ be the reminder when we divide $$10^{s}n=\overline{b_1b_2\cdots b_s.[R][R]\cdots}$$ by $d$. Since $\gcd(10^sn,d)=1$, we see that $\gcd(r_0,d)=1$ with $0<r_0<d$.
Since $\dfrac{r_0}{d}=\overline{0.[R][R]\cdots}$, we get $\dfrac{r_0}{d}=\dfrac{\overline{r_1r_2\cdots r_t}}{10^t-1}$.
It follows from this that $d$ is a divisor of $r_0(10^t-1)$. Since $\gcd(d,r_0)=1$, we see that $d$ is a divisor of $10^t-1$.
Let $u$ be the smallest positive integer $x$ such that $10^x\equiv 1\pmod d$.
Then, $u$ is a divisor of $t$. (The reason is as follows : There are non-negative integers $y,r$ such that $t=uy+r$ and $0\le r\lt u$, so $1\equiv 10^t\equiv 10^{uy+r}\equiv (10^u)^y\cdot 10^r\equiv 10^r\pmod{d}$. Now, $r\gt 0$ contradicts that $u$ is the smallest positive integer such that $10^x\equiv 1\pmod d$. So, we have $r=0$ which implies that $u$ is a divisor of $t$.)
Since $10^u-1$ is a multiple of $d$, we see that $n(10^u-1)$ is also a multiple of $d$.
Therefore, $\dfrac{n(10^u-1)}{d}$ is an integer.
So, $0\lt \dfrac nd\lt 1$ implies $0\lt \dfrac{n(10^u-1)}{d}\lt 10^u-1$. So, $\dfrac{n(10^u-1)}{d}$ is a positive integer, and the number of the digits is at most $u$.
Letting $\dfrac{n(10^u-1)}{d}=\overline{c_1c_2\cdots c_u}$, we have $$\dfrac{n}{d}=\dfrac{\overline{c_1c_2\cdots c_u}}{10^u-1}=\overline{0.[R'][R']\cdots}$$
where $[R']=\overline{c_1c_2\cdots c_u}$.
This means that the repeating digits appear right after the decimal point.$\ \blacksquare$
