The Weierstrass $\sigma$ function of the lattice $\Lambda$ is defined by $$\sigma (z;\Lambda)=z\prod_{\lambda\in\Lambda;\,\lambda\ne 0}\left(1-\frac{z}{\lambda}\right)\exp\left(\frac{z}{\lambda}+\frac{1}{2}\left(\frac{z}{\lambda}\right)^2\right).$$ Is the factor $\exp\left(\frac{z}{\lambda}\right)$ extraneous? I.e. is this equivalent to $$\sigma (z;\Lambda)=z\prod_{\lambda\in\Lambda;\,\lambda\ne 0}\left(1-\frac{z}{\lambda}\right)\exp\left(\frac{1}{2}\left(\frac{z}{\lambda}\right)^2\right)?$$
For every lattice point $\lambda$ there is a lattice point $-\lambda$ on the same lattice and they cancel each other: $$\exp\left(\frac{z}{\lambda}\right)\exp\left(\frac{z}{-\lambda}\right)=1$$
Is that reasoning correct? If not, where is the error?
You are right that for every $\lambda\in\Lambda$ there is a corresponding $-\lambda\in\Lambda$. Thus, if $\sum 1/\lambda$ converged, it would converge to $0$. But, it only converges conditionally. To think about the convergence of products, we typically take logarithms and think about the convergence of series. In this case, $\ln(1-z/\lambda)=(-1+o(1))z/\lambda$. Compare to the 1D version: the harmonic series + its negative. Of course, any of the "most obvious" ways to sum the series would do so in a symmetric manner, such as ordering terms according to magnitude, and thus yield $0$, but for bookkeeping purposes it's generally a good idea to stick to absolutely convergent things. IIRC, this isn't strictly necessary (I think some sources on Weierstrass elliptic functions live with conditional convergence), but it's a common and good idea which the source you're reading is using.
The culprit is that $(1-z/\lambda)\to1$ too slowly. To mitigate $(1-z/\lambda)$'s slow convergence, we need "offset factors" whose partial products mirror that of the original factors, i.e. converge to $1$ "just as slowly" in a fashion which "mirrors" that of $(1-z/\lambda)$ (e.g. the offset factor is "above" $1$ when the other is "below"). It is easiest to see the necessary cancellation by using $\ln$s again: the offset factors' logarithm must grow as $(+1+o(1))z/\lambda$, which means $\exp(z/\lambda)$ is the perfect choice for an offset factor.
This kind of reasoning is behind the general form of all Weierstrass factorizations. Note the elementary factors $E_n(z)=(1-z)\exp(\Sigma^n_{k=1}z^k/k)$, when we take the logarithm, are $\ln(1-z)$ plus the $n$th partial sum of the Newton-Mercator series for $-\ln(1-z)$.