Tail Sum Formula states that:
For $X$ with possible values $\{0, 1, 2, \ldots , n\}$, $$\operatorname E(X) = \sum_{j=1}^n P(X \ge j)$$
Notice the $j$ condition starts at $1$ not $0$ because $E(X)=\sum_{x=0}^{n}xP(X=x) = \sum_{x=1}^{n}xP(X=x) $
Then that means for $X \in \{1, 2, ..., n\}$
$$E(X) = \sum_{j=1}^n P(X \ge j)$$
The $X$s have different ranges, but have equal $E(X)$. How could this be?
The fact that you do not use the $0$ case in your calculation does not change the range of $X$. We still have $X$ taking on the value of $0$ with probability $P(X=0)$, this just does not count towards the expectation.
Here is a simple counterexample to your range claim. Take $X$ with $P(X=0)=1/2$ and $P(X=1)=1/2$. I hope that we can agree this has $X$ expectation $1/2$. But now if we restrict the range of $X$, the probability distribution does not make sense anymore, as we must have probabilities sum to one and all we have is $P(X=1)=1/2$ and $P(X \neq 1)=0$. Even if you tried to account for this by scaling probabilities so $P(X=1)=1$, we now get $1$ as the expectation, a different value than we had before.
Also, it may be useful to point out that the two expressions are algebraically equivalent, that is, $$\operatorname E(X) = \sum_{j=1}^n P(X \ge j)=\sum_{j=1}^n \sum_{i=j}^nP(X=i)=\sum_{i=1}^{n}iP(X=i)$$
Hopefully that clears up any confusion.