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.)

First off, if you're doing an indefinite integral, $I$ is a function, so let's write $I(x)$. Second, as pointed out, once we make a substitution in your second example, we no longer have the same function $I()$.

So in your second example, under the overall assumption that $u=-x$ and being stricter about functions, you'd be making the argument

$$I(x) = .... = J(u) = J(-x)$$

where $J()$ is the name introduced for the new function we have after substitution.

Now in some cases we do have $J(-x) = I(x)$ and we can do the "bring over to the other side" trick and get a final answer for $I$. But in the cases where the trick goes bad we can see that $J(-x)$ is not equal to $I(x)$ , and the trick wasn't permitted in the first place.