The problem is:
Give an example of a convergent series such that, when the terms are rearranged, the series sums to a different value.
A solution is:
Although everything makes sense in this solution, I don't get how what it claims is possible. How can a series have two different sums?
Is this caused by the fact that subtraction is not commutative or associative? Hence, in fact, we really have two different series here?

Recall that the sum of the series is the limit of partial sums.
Now, by reordering the sum, you change the partial sum, and hence you get a different sequence, which can converge to a different limit.
This is actually what happens with conditionally convergent series, but NOT with absolutely convergent ones.
To understand better what happens, consider a series $\sum_n a_n$. Define $$b_n=\left\{ \begin{array}{lr} a_n & \mbox{ if } a_n \geq 0 \\ 0 & \mbox{ if } a_n < 0 \\ \end{array} \right. \\ c_n=\left\{ \begin{array}{lr} a_n & \mbox{ if } a_n < 0 \\ 0 & \mbox{ if } a_n \geq 0 \\ \end{array} \right. \\$$ Then, if $\sum a_n$ is conditionally convergent, then $\sum b_n=\infty, \sum c_n = -\infty$. If we denote the partial sums by $S_, T_n$ and $R_n$ respectively then $$S_n=T_n-R_n$$
Now, by reordering, since both $b_n$ and $c_n$ take the zero value infinitely many times (again by conditinal convergence) you get new partial sums $$S_n'=T_n'-R_n'$$
Since $\sum_{n} b_n =\infty$ you can make by rearangement $T'_n$ as big as you want. Also since $b_n=0$ infinitely many times, you can also make $T'n$ very small compated with $n$.
Same way, you can make $R_n'$ as big or as small as you want. Therefore, you can make the difference $S_n'$ anything you want
To make this more clear:
Pick your favorite number $a$. Rearrange the conditionally convergent series $a_n$ the following way:
Start adding the positive terms ($b_n>0$) until the first time you exceed $a$. Then add negative terms until the first time you go below $a$. Then add the following positive terms until the first time you exceed $a$, then negative ....
Since $\sum b_n=\infty, \sum c_n = -\infty$, by the rearrangement you get a series that oscillates around $a$ infinitely many times. Since by the so-called divergence test you get $\lim_n a_n=0$ the new series oscillates around $4a$ by bounds which converge to $0$, hence it is convergent to $a$.
But the number $a$ was chosen arbitrarily.
Again, the point is that by moving more positive or negative terms at the beginning of the series you get larger or smaller partial sums, which then can converge to a different limit.