Is there an extension of the integers where the "sum of natural numbers" is rigorous?

There's the well-known claim that $$\sum_{n=1}^{\infty} n = -\frac{1}{12} \tag 1$$ Of course in this form, using the usual interpretation of the infinite sum as limit of finite sums, it's wrong, as the sum on the left hand side diverges. The $-1/12$ is obtained by using the series of the zeta function, $$\zeta(s) = \sum_{n=1}^{\infty} n^{-s}$$ and then observing that formally evaluating it for $s=-1$ gives eq. (1).

Now let's consider another "sum" of a divergent series: $$\sum_{n=0}^{\infty} 2^n = -1 \tag 2$$ Again, you can do an argument like the above: If we take the geometric series $$\frac{1}{1-q} = \sum_{n=0}^{\infty} q^n$$ and insert $q=2$, we get the claimed identity.

However, eq. (2) can also be made rigorous by extending the integers to the $2$-adic numbers, where the left hand side indeed converges to $-1$.

My question therefore is:

Does there exist an extension of the integers that makes eq. (1) rigorous in the sense that in that extension the sum actually converges to the value $-1/12$?

The $2$-adic topology makes $\mathbb{Z}$ into a Hausdorff topological group (and in fact a topological ring). This is important: the fact that the topology is compatible with addition is used to prove several basic facts about convergent series, e.g. that changing a finite number of terms does not affect convergence.

Claim: There is no Hausdorff topological group structure on $\mathbb{Z}$ such that the sequence $$ \tag{$\star$} \sum_{n=1}^kn,\;\;\;k=1,2,\ldots $$ is Cauchy.

Proof: Suppose such a topology exists. Choose an open neighborhood $U$ of $0$ such that $1\not\in U$, and a neighborhood $V$ of $0$ such that $V-V\subset U$. Here, $V-V=\{v_1-v_2:v_1,v_2\in V\}$. We assume $(\star)$ is Cauchy, so there is some $N>0$ such that $$ \sum_{n=1}^{k_1}n-\sum_{n=1}^{k_2}n\in V $$ whenever $k_1$, $k_2>N$. Now $$ 1=\left(\sum_{n=1}^{N+3}n-\sum_{n=1}^{N+2}n\right)-\left(\sum_{n=1}^{N+2}n-\sum_{n=1}^{N+1}n\right)\in V-V\subset U, $$ which contradicts $1\not\in U$.