Yes but you are doing a fractional algebra. To refresh your memory:
a/b + c/d = (a*d + c*b)/(b*d)
Since numerator stays b*d, you maintain the same poles. However your denominator is changed significanty, meaning you do not necessarily maintain the same zeros after the addition.
In transfer function terms
(s-2)/(s-5) + (s-1)/(s-3) = [(s-2)*(s-3) + (s-1)*(s-5)] / [(s-5)(s-3)]
So your numerator is the same, poles of your tf1 and tf 2. However your denominator that defines your zeros is now
(s-2)*(s-3) + (s-1)*(s-5) = 2*s^2 - 11*s +11
So zeros of your tf1 and tf2 are no longer the zeros when you add them together
