Best case analysis for Shell sort
The best case for Shell sort is when the array is already sorted.
In that case the inner loop does not need to any work and a simple comparison will be sufficient to skip the inner sort loop.  

A best case of $Θ(N)$ is reached by using a constant number of increments.
But the problem is that the best value for the gapping depends on the input. Therefore you cannot choose a single best value for the gap and must suffer a $log^2(N)$ delay due to repeated compares.    

So you are almost correct. Obviously there is no way to compare fewer than N-1 elements before we can conclude that the array is already sorted, so $Θ(N\space log^2(N) )$ it is.  

Note that this best case is for randomized Shell sort which is subtly different from the original.  

See: https://arxiv.org/pdf/0909.1037.pdf

All this means that Shell sort is not a particularly good algorithm for a (nearly) sorted array. The gapping introduces redundant compares whereas other sorting algorithms only have to go over the array once in the best case.
When the array is random, the gapping works in your favor, and these other algorithms that perform so wonderfully on a (nearly) sorted array perform at a dismal $O(N^2)$