What is the time complexity of a binary multiplication using Karatsuba Algorithm?
You're mistaken.
The analysis of the Karatsuba algorithm takes addition and subtraction into account.

If you analyze your pseudo-code, you can see that you have exactly 3 recursive function calls with arguments of size $n/2$, and all other operations like addition, subtraction, extracting the higher and lower bits, ...,  run in $O(n)$.

Therefore you get the recursion $T(n) = 3 \cdot T(n/2) + O(n)$ for the complexity, which gives the following according to the Master theorem: $$T(n) = \Theta(n^{\log_2 3}) = \Theta(n^{1.58...})$$



Fun fact: even if you assume that all the bitwise operations (addition, subtraction, ...) take $O(1)$, the code will still have the same complexity.
Since the number of calls to the three subproblems dominates the other work. But I doubt, that any book/lecturer would assume that the bitwise operations are constant, since the whole point of the algorithm is that you want to multiply large numbers.



Btw, you have some small errors in your pseudo-code/analyze.
E.g. here:

long resA = a * (long) Math.pow(10, (2*halfMax)); // takes O(1)


You want to multiply with a power of 2, to make it efficient (since you are working with binary numbers). And even corrected it will not run in $O(1)$, because you need to copy/move $O(n)$ bits. So this line will also run in $O(n)$.