(This elaborate answer is meant to thoroughly justify why your proof is wrong, and to give you a little more insight into why it is.)
The answer is no.
It is a result due to Riemann that the suitable "infinite extension" of the binary operation of addition (with all the field rules that you mention in your faulty proof) is not merely the concept of series, or convergent series, but that of absolutely convergent series, meaning that you need require that both
$$\sum_{n\in\mathbb N} a_n \qquad \mathrm{and} \qquad \sum_{n\in\mathbb N} |a_n| $$
be convergent series. The main issue of series that do not absolutely converge concerns the possibility of reordering the terms (the "extension" of the commutative property that you wanted to use early on in the proof - actually, in the case of divergent series, associativity becomes a problem too, and Grandi's series is the common example). This is true of any kind of series – real, complex, vector-valued – although it is simpler to only deal with the real case.
More precisely, Riemann showed that:
If one reorders the terms of an absolutely convergent series, the new series is still absolutely convergent and will moreover converge to the same value as the original one;
Instead, given any real number $\lambda$ (!!), it is possible to reorder the terms of a conditionally (i.e. not absolutely) convergent series such that the new series converges to $\lambda$, or such that it oscillates, or even diverges.
I'll give you an example to convince you of these facts.
Oscillating rearrangement of a conditionally convergent series. Let $\sum_{n=0}^\infty a_n$ be a conditionally convergent series: we construct an oscillating rearrangement. Choose ad libitum two real numbers $\alpha,\beta$ with $\alpha < \beta$: our rearrangement will have to be such that, when you inspect the sequence
$$A_p \doteq \sum_{n=0}^p a_n$$
you will always be able to select an infinite amount of terms that lie below $\beta$ and an infinite amount of them that lie above $\alpha$. Heuristically, this means that as $p$ grows, the sequence $A_p$ goes up above $\alpha$, then goes down below $\beta$, and continues to do this ad infinitum.
I introduce now the series $\sum_{n=0}^\infty (a_n)^+$ and $\sum_{n=0}^\infty (a_n)^-$, respectively the series of the positive terms of $a_n$ and the series of the negative terms of $a_n$. For example, if your series is the alternating harmonic series
$$\sum_{n=1}^\infty b_n = 1- \frac 1 2 + \frac 1 3 - \frac 1 4 + \cdots = \ln 2 $$
where $b_n = \frac{(-1)^{n+1}}{n}$, then
$$\sum_{n=1}^\infty (b_n)^+ = 1 + \frac 1 3 + \frac 1 5 + \cdots \qquad \mathrm{and} \qquad \sum_{n=1}^\infty (b_n)^- = - \frac 1 2 - \frac 1 4 - \frac 1 6 - \cdots $$
The fundamental fact we exploit in our construction is that
Lemma. If $\sum_{n=0}^\infty a_n$ is conditionally convergent, then $\sum_{n=0}^\infty (a_n)^\pm$ diverge.
(Check this for the series above!)
What does the last statement mean? It means that if I denote, as before,
$$B_p \doteq \sum_{n=0}^p (a_n)^+ \qquad \mathrm{and} \qquad C_p \doteq \sum_{n=0}^p (a_n)^- $$
then, if you gave me two positive numbers $M$ and $N$, however large you want them to be, I will always be able to find two indices $p'$ and $p''$ such that
$$\forall p > p' \quad B_p > M \qquad \mathrm{and} \qquad \forall p > p'' \quad C_p > N$$
Now, we can suppose w.l.o.g. that $a_0 < \beta$ and $a_n \neq 0$ for all $n$. Call $n_0 \doteq 0$ and consider the first index $n > n_0$ such that $a_n > 0$: call it $n_1$. Consider the sum $a_{n_0} + a_{n_1}$: if it is greater than or equal to $\beta$, we stop. Otherwise, let $n_2$ be the first index $n > n_1$ such that $a_n > 0$, consider the sum $a_{n_0} + a_{n_1} + a_{n_2}$ and compare it to $\beta$: if it is greater than or equal to $\beta$ we stop, otherwise we keep on adding positive contributes to our sum. Notice that these positive contributes will always exist, because $\sum_{n=0}^\infty (a_n)^+$ diverges; for the same reason, we only need a finite number of steps to get to the point where our sum $S'$ is greater than or equal to $\beta$. Indeed, when we get there, we will have selected say $m$ indices $n_0, n_1, \dots, n_m$, that are finitely many and are all distinct, every one of which correspond to the first $m$ terms of our rearrangement: call them $b_0,b_1,\dots,b_m$. (Also notice the sum $S'$ of these first $m$ terms is the first of the infinitely many terms of the sequence $A_p$ that we can choose that lie above $\beta$!)
Now we start again from the first values of $n$ and we try to fall below $\alpha$ by adding negative terms. Consider the first index $n$ such that $a_n < 0$: call it $n_{m+1}$ and inspect the sum $S' + a_{n_{p+1}}$. If it is less than or equal to $\alpha$ we stop. Otherwise we select the first index $n > n_{p+1}$ such that $a_n < 0$, we call it $n_{p+2}$ and we inspect the sum $S' + a_{n_{p+1}} + a_{n_{p+2}}$. If this is less than or equal to $\alpha$ we stop; otherwise (you get the jist) we keep on adding negative terms until we reach the sum $S'' \leq \alpha$. This we can do in a finite number of steps because $\sum_{n=0}^\infty (a_n)^-$, too, diverges; call the last index we needed to find $n_q$ and rename the terms of the sequence $a_n$ corresponding to $n_{p+1},n_{p+2},\dots,n_q$ as $b_{p+1},b_{p+2},\dots,b_q$. As before, the number $S''$ is the first element of the subsequence of $A_p$ that lies completely below $\alpha$.
Observe that, even if we suppress the (still finitely many) terms we've already selected from the two series $\sum_{n=0}^\infty (a_n)^\pm$, what remains is still a pair of divergent series, so we may iterate this procedure indefinitely.
And there is you oscillating rearrangement! You may try to follow the procedure for the alternating harmonic series choosing e.g. $\alpha = -1$ and $\beta = \pi$. (There are similar procedures to build divergent rearrangements, or rearrangements that converge to a fixed real number $\lambda$.)
