Let $a\in \mathbb{Q}$ such that $18a$ and $25a$ are integers, then we wish to prove that $a$ must be an integer itself. What that means is that $a=\frac{p}{1}$ where $p \in \mathbb{Z}$. What we do know is that we can express the $\gcd(18,25)$ as: $$ \gcd(18,25)=18x +25y$$ Now if $x=y=a$, we are done, since: $$ \gcd(18,25)=18a +25a=43a$$ as the $\gcd$ is always an integer and so is 43, so $a$ is also an integer.
But, how would I generalise this?
On
You can also prove this by contradiction:
Assume $a=\frac{m}{n}$, where $m,n$ are coprime and $n\gt1$. Then, if $18a=k_1\in\mathbb{Z}$, by the fundamental theorem of arithmetic $$n=2^b3^c.$$ However, assuming $$25a=k_2\in\mathbb{Z},$$ implies $n=5$ or $n=25$, which is obviously a contradiction.
On
Another way to look at this: $18a$ and $25a$ are integers. Therefore, so is $25a-18a = 7a$.
Therefore so is $18a-2(7a) = 4a.$
Therefore so is $7a-4a = 3a.$
Therefore so is $4a-3a = a.$
On
Conceptually $\, \dfrac{m}{18} \!=\! a \!=\! \dfrac{n}{25}\,$ so its least denominator divides coprimes $18,25$ so is $1,\,$ so $\,a\in\Bbb Z$
Remark $\ $ This is an additive analog of this better-known (multiplicative) group result
$$ a^{\large 18} = 1 = a^{\large 25}\,\Rightarrow\, {\rm ord}(a)\mid 18,25\,\Rightarrow\, {\rm ord}(a)=1\,\Rightarrow\, a = 1\qquad$$
Generally if a rational can be written with denominators $\,a\,$ and $\,b\,$ then it can be written with denominator $\,\gcd(a,b).\,$ See also here and see here for four proofs and elaboration.
For further discussion see denominator ideals and order ideals and unique fractionization.
All you know is that there are some $x$ and $y$ with that property, but that doesn't imply that you can take $x=y=a$.
Note that $\gcd(18,25)=1$. Therefore, there are integers $x$ and $y$ such that $18x+25y=1$. But then $a=18xa+25ya\in\mathbb Z$, since $18a,25a,x,y\in\mathbb Z$.