If you're doing an indefinite integral, $I$ is a function. In your first example, it is always a function of $x$, so no problem arises. But in your second example you'd get
$$I(x) = .... = I(u) = I(-x)$$
and you can't bring the $I$ back to the other side and get $2I$.
(Confession: I haven't fully thought this through, so I'm not sure it's a 100% valid explanation of what's happening. But $I$ being a function is definitely an issue - I'm just not sure if it's pointing to a problem with the argument, or just notation, or a combination of the two.)