Does there exists a finite abelian group $G$ containing exactly $60$ elements of order $2$?

Yes, your understanding is correct.

Consider the relation $\sim$ on $G$ (having even order, otherwise it has no element of order $2$) defined by $$ a\sim b \quad\text{if and only if}\quad (b=a\text{ or }b=a^{-1}) $$ This is easily seen to be an equivalence relation. The equivalence classes have either one or two elements. If you remove the two-element equivalence classes, you are dropping an even number of elements from $G$, so what remains is an even number. Drop also the class consisting of $1$ and you remain with an odd number of one-element equivalence classes: these elements are precisely those having order $2$.

Clearly, the order of $G$ cannot be an odd, it's obvious. Suppose the order of $G$ is $2n$. Since order of identity is always $1$ i.e. $|e|=1$. So we have left only $2n-1$ element which is odd in number. Elements which are not their own inverses, these elements and their inverses exist in pairs.

Then it should be even in number but we have odd number here. In that case we must have atleast an element which must be self inverse. Elements of order $2$ are exactly those elements which are self inverse. Thus number of element of order $2$ must be in odd number. But we have given $60$ number of elements of $2$ which is absurd.