Can we easily check whether a number is in this sequence? Here :
https://oeis.org/A060650
the positive integers $\ n\ $ are shown for which there is exactly one non-abelian group of order $\ n\ $.

Is there an easy criterion to check whether a number is in this sequence ?

It seems that a number in this sequence must be cubefree (Is this actually true ?) I am also aware of a formula for the number of groups of order $\ n\ $ when $\ n\ $ is squarefree, although this formula is quite complicated.
If the number is even, it seems that it is in the list if and only if it is twice an odd prime.
 A: Let $n$ be cubefree and call a prime $p$ dividing $n$ "simple" if $p^2$ does not divide $n$, and "double" otherwise. As we run over the pairs of primes dividing $n$, $p$ and $q$, count the following occurences.
A. $p\mid q-1$ and $p$ or $q$ is a double prime.
B. $p\mid q-1$ and both $p$ and $q$ are simple primes. 
C. $p\mid q+1$ and $p$ and $q$ are double primes.
D. $p\mid q+1$ and $p$ is a simple prime and $q$ is a double prime.
Then I think $n$ is in your list if and only A and C occur $0$ times, while B and D occur exactly once in combined total.
Here's a sketch of a proof. First, if $4$ divides $n$, then $n$ is not in the list. (If $n=4$, then $G$ is abelian, otherwise we can easily find two non-abelian groups, say dicyclic, and dihedral direct with cyclic.) So we may assume $4$ does not divide $n$ and $n$ is soluble, so all the Hall subgroups exist.
If all the Hall subgroups involving two primes are abelian, then I think $G$ is abelian, so we need some Hall subgroup involving two primes to be nonabelian. This is about groups of order dividing $p^2q^2$ and I think A, B, C and D are the only possibilities.
Now, suppose $p\mid q-1$ and $q$ is a double prime. Then we get two nonabelian groups (say $q\times (q\rtimes p)$ and $(q^2\rtimes p$). If $p$ is a double prime, then we have $p\times (q\rtimes p)$ and $(q\rtimes p^2)$. We can use this to create different nonabelian groups $G$. This shows A cannot occur.
Similarly, if C occurs, we get at least two nonabelian groups, by starting with $(q^2\rtimes p)\times p$ and $q^2\rtimes p^2$, so C cannot occur.
Finally, if B or D occur, we get a nonabelian group, and if we get more than one occurrence, it's not hard to show that there are at least two different nonabelian groups.  (It remains to show that if indeed there is only one occurrence, then we get only one nonabelian group.)
