Which functions are tempered distributions?

Question 1 What you have is almost enough. Assume $\Re\sigma \leq -\epsilon < 0$. Test $\exp (-\sigma |x|^2)$ "against" $\phi(x) = \exp( (\sigma+\epsilon/2)|x|^2)$ in the following way: you can construct a sequence of annular cut-off functions $\chi_k$ such that $\chi_k \phi \to 0$ in $\mathcal{S}$ (using the exponential decay of $\phi$) and $\langle g_\sigma, \chi_k\phi\rangle > c > 0$ for all $k$.

Question 2 You have the structure theorem of tempered distributions. (See Theorem 8.3.1 in Friedlander and Joshi).

Theorem Every tempered distribution is a (distributional) derivative of finite order of some continuous function of polynomial growth.

If you intersect against $L^1_{loc}$, this just guarantees that the distributional derivative is actually the weak derivative. From this you can conclude that an appropriate version of what you stated is true.

Thinking $L^1_{loc}\cap\mathcal{S}'$ as a subset of $\mathcal{D}'$, it is not true that every element of $L^1_{loc}\cap\mathcal{S}'$ is a polynomially growing function.

For example, in $\mathbb{R}$, define

$$f:\mathbb{R}\to\mathbb{R}, t\mapsto \cos(e^t) e^t.$$

Then $f\in L^1_{loc}(\mathbb{R})$, so it represents by integral pairing an element of $\mathcal{D}'(\mathbb{R})$.

Also, define:

$$g:\mathbb{R}\to\mathbb{R}, t\mapsto \sin(e^t).$$

Then $g\in L^\infty(\mathbb{R})$, so it represents by integral pairing an element of $\mathcal{S}'(\mathbb{R})$.

Also, denoting the distributional derivative with the symbol $D$, we have that:

$$\forall\varphi\in\mathcal{D}(\mathbb{R}), f(\varphi)= \int_\mathbb{R}f(t)\varphi(t)\operatorname{d}t =\int_\mathbb{R} \cos(e^t) e^t\varphi(t)\operatorname{d}t \\ = \int_\mathbb{R} \left(\frac{\operatorname{d}}{\operatorname{d}t}\sin(e^t)\right)\varphi(t)\operatorname{d}t =- \int_\mathbb{R} \sin(e^t)\varphi'(t)\operatorname{d}t = -g(\varphi')=Dg(\varphi).$$ So, being $\mathcal{S}'(\mathbb{R})$ closed with respect to distributional derivative, we get that $f=Dg\in\mathcal{S}'(\mathbb{R})$.

However $f$ is not a polynomially growing function, so we have got an example of $f\in L^1_{loc}(\mathbb{R})\cap\mathcal{S}'(\mathbb{R})$ that is not of polynomial growth.