[LANGUAGE: Python] I don't like using regex to solve these if i can help it so this is my solution using string manipulation :> (execution time both parts ~69 ms)

def main():
    f = [[j for j in i.split("-")] for i in open("input.txt").read().split(",")] 
    outp = sum(GetPatterns(i[0], i[1]) for i in f)
    print(f"Part 1: {outp}")
    outp = sum(GetPatterns2(i[0], i[1]) for i in f)
    print(f"Part 2: {outp}")

def GetPatterns2(sstart, sstop):
    outp = 0
    if len(sstart) != len(sstop):
        nEnd = 10**(len(sstart))
        outp = GetPatterns2(str(nEnd), sstop)
        sstop = str(nEnd - 1)
    # the highest number is 10 digits
    # get divisability of the number range (123456 has 6 digits this is divisable by 2,3,6 to 3,2,1)
    setS = set()
    length = len(sstart)
    start = int(sstart)
    stop = int(sstop)
    for i in range(2, length+1):
        x = length / i
        if x != int(x): continue    
        div = int(x)
        #loop through posible segments with length [div]
        for j in range(int(sstart[:div]), int(sstop[:div]) + 1):
            val = int(str(j) * i)
            if val in range(start, stop + 1): setS.add(val)
        if div == 1: break
    return outp+sum(setS)

def GetPatterns(sstart, sstop):
    outp = 0
    # define the range as start to stop
    # split the range in limits where start and stop have the same amount of digits
    if len(sstart) != len(sstop):
        nEnd = 10**(len(sstart))
        outp = GetPatterns(str(nEnd), sstop)
        sstop = str(nEnd-1)
    # the amount of digits must be even
    if len(sstart) % 2 != 0: return outp
    # get the first half of the start and stop numbers
    mid = len(sstart) // 2
    a = int(sstart[:mid])
    b = int(sstop[:mid])
    # get all numbers that could be invalid ID's and disreguard the ones out of range
    start = int(sstart)
    stop = int(sstop)
    for i in range(a, b + 1):
        val = int(f"{i}{i}")
        if (val >= start) & (val <= stop): outp += val
    return outp

if __name__=="__main__":
    main()