$e-2\approx0.71828$, but I got $1$
Consider the numerators of the negative terms. The first one, call it $a_3$ for ease of notation, is $2$. Then they follow the recursion $a_{n+1}=(n+1)a_n-1$. What's the behavior of the sequence $\frac{a_n}{n!}$? Cases:
- It doesn't go to zero. In this case, your series doesn't even converge.
- It goes to zero but isn't summable. In this case, Riemann's theorem on conditional vs. absolute convergence will warn you that such a rearrangement may (or may not) change the value of the sum, or even cause it to fail to converge at all.
- It is summable. In this case such a rearrangement cannot change the value of the sum.
Also note that you are only considering the sequence of partial sums at even indices (i.e. the sum of 2 terms, 4 terms, ...). This can cause a sum that doesn't converge to look like it does; consider for a simpler example $\sum_{n=0}^\infty (-1)^n$.
You can do this with any series.
You have a sum $a_1+a_2+a_3+a_4+\cdots$.
You can rewrite it as $1+(a_1-1)-(a_1-1)+(a_2+a_1-1)-(a_2+a_1-1)+(a_3+a_2+a_1-1)-\cdots$
The partial sums are $1, a_1, 1, a_2+a_1, 1, a_3+a_2+a_1, \dots$
Clearly splitting the elements in this way does not help any to sum the original series.
In response to the comments what OP has done in the question is rather split as $$a_1+(1-a_1)-(1-a_1-a_2)+(1-a_1-a_2)-\dots$$ with partial sums $a_1, 1, a_1+a_2, 1 \dots$
Whichever is used the "adjustment" term is of the form $\pm \left(1-\sum_{r=1}^n a_r\right)$. Here it tends in absolute value to $1-(e-2)=3-e$.
You are only shifting the difference at the infinity!
EG
note that $$1-\frac {203}{6!}=1-\frac {203}{720}=0.71805\ldots$$
More in general, it can be easily shown that the remainder is solution of the following recurrence equation:
$$a_k=a_{k-1}-\frac{1}{k!}$$
$$a_3=\frac13$$
which quickly converges to the value: 0.281718... = 1-e
https://www.wolframalpha.com/input/?i=a(k)%3Da(k-1)-1%2Fk!,+a(3)%3D1%2F3