How does squaring time complexity imply the same time complexity for multiplying different numbers? Isn't it the other way round? Found this in solutions of a test as being true

If you can square an n-bit integer in time $O(n \,log \,n)$, then you can multiply two n-bit integers in time $O(n \, log \,n)$.

How does the above scenario apply for multiplying different numbers? Isn't it possible that squaring is a special operation since it is just 1 number? . I would see this making sense.

If you can multiply two n-bit integers in time $O(n \, log \,n)$, then you can square an n-bit integer in time $O(n \,log \,n)$.
Of course, we can suppose we are able to do sum and difference in linear time. Now, if we want to calculate $h\cdot k$, we can find in linear time $a$ and $b$ such that $h=a-b$ and $k=a+b$: take $a=\frac{k+h}{2}$ and $b=\frac{k-h}{2}$. Then $h\cdot k=(a-b)(a+b)=a^2-b^2$, so we only need to square two numbers of at most $n$ digits (if $h$ and $k$ have $n$ digits, then also $\frac{k+h}{2}$ has at most $n$ digits) in $O(n\log(n))$ and take their difference in $O(n)$.
Actually, this works if $h$ and $k$ have the same parity, else $a$ and $b$ are not integers; if it is not the case, then you can write $h\cdot k=h\cdot(k-1)+h$ and proceed as above.