I have my own solution that I think is way less complex and similar efficiency.

We have outer loop which is O(N) and then we do an inner loop to get subarrays and we unpack arrays as we add them making the overall time complexity O(N^3)

Space O(N^3)

class Solution:
  def findSubarrays(self, arr, target):
    result = []
    end = len(arr)
    # TODO: Write your code here
    # Go through the entire array
    for i, num in enumerate(arr):
      # Pointer starts after current index (for continuous array)
      pointer = i+1
      product = num
      product_arr = [num]
      # Technically it is possible to have a current number that is more than target
      # But 0 as the next number - this would mean individual target is not but joined it is
      if product < target:
        result.append(product_arr)
      while pointer < end and arr[pointer]*product < target:
        # If we are still less than target, append a new array to result and update product!
        product *= arr[pointer]
        product_arr=[arr[pointer], *product_arr]
        result.append(product_arr)
        pointer += 1

    return result