The answers are yes and no.
Yes. Denote by $g$ the first of mentioned generators of order $3$. Consider the embedding $E\subset{\mathbb P}_k^2$ given by $\mathcal L$, pick as $0\in E$ one of the inflection points, and denote by $\oplus$ the operation on $E$. Let $p_1+p_2+p_3$ be the divisor of the intersection of $E$ with the line $z_i=0$. Then $p_1\oplus p_2\oplus p_3=0$. By construction, $p_1+p_2+p_3=(p_1\oplus g)+(p_2\oplus g)+(p_3\oplus g)$, implying that the $p_i$'s are of order $3$. Hence, all $9$ inflection points lie on the "parallel" lines $z_i=0$ and, therefore, provide a Hesse pencil (see http://en.wikipedia.org/wiki/Hesse_configuration and http://en.wikipedia.org/wiki/Hesse_pencil for details).
No. As line bundles of degree $3$ up to an isomorphism are divisors of degree $3$ up to linear equivalence, the moduli space of such line bundles is isomorphic to the proper elliptic curve $E$. At the level of divisors, every divisor $D$ on $E$ of degree $3$ is equivalent to $2\cdot0+p$, $p\in E$, so we map $[D]\mapsto p$. On the other hand, there is a map $j:{\mathbb P}_k^1\to{\mathbb P}_k^1$ of degree $9$ (first ${\mathbb P}_k^1$ is the Hesse pencil) that calculates the $j$-invariant of $E$ (see http://arxiv.org/pdf/math/0611590v3.pdf). Which of the $9$ possible points of the Hesse pencil you get depends on the choices you made while constructing the pencil and not on the choice of ${\mathcal L}$"$\in$"$E$.