Rounding a number in Python but keeping ending zeros

As of Python 3.6, you can also use an f-string to inline format the number. In this case, the desired format is floating point with 2 decimal places so you would use .2f as the format specifier:

x = 2606.89579999999
x = round(x, 2)      # not strictly necessary as format will round for you
print(f'{x:.2f}')

Output:

2606.90

As you are talking about trailing zeros, this is a question about representation as string, you can use

>>> "%.2f" % round(2606.89579999999, 2)
'2606.90'

Or use modern style with format function:

>>> '{:.2f}'.format(round(2606.89579999999, 2))
'2606.90'

and remove point with replace or translate (_ refers to result of previous command in python console):

>>> _.translate(None, '.')
'260690'

Note that rounding is not needed here, as .2f format applyies the same rounding:

>>> "%.2f" % 2606.89579999999
'2606.90'

But as you mentioned excel, you probably would opt to roll your own rounding function, or use decimal, as float.round can lead to strange results due to float representation:

>>> round(2.675, 2)
2.67
>>> round(2606.89579999999, 2)
2606.89

With decimal use quantize:

>>> from decimal import *
>>> x = Decimal('2606.8950000000001')
# Decimal('2606.8950000000001')
>>> '{}'.format(x.quantize(Decimal('.01'), rounding=ROUND_HALF_EVEN))
'2606.90'

That, for your original task, becomes:

>>> x = Decimal('2606.8950000000001')
>>> int((x*100).quantize(1, rounding=ROUND_HALF_EVEN))
260690

And the reason of strange rounding comes to the front with Decimal:

>>> x = Decimal(2606.8950000000001)
# Decimal('2606.89499999999998181010596454143524169921875') # internal float repr