Every short exact sequence of groups induces an action of the quotient on the normal subgroup by conjugation. Here, the action is inversion in $C_3$; to see this explicitly, observe that $$(12)(123)(12)^{-1} = (132).$$ Anyway, you want to know what the map $i^*: H^*(C_3; \Bbb Z) \to H^*(C_3;\Bbb Z)$ induced by inversion is. The first useful fact is that because inversion is a group homomorphism, this map is an algebra homomorphism.
Because $H^*(C_3;\Bbb Z) = \Bbb Z[c]/(3c)$, where $|c| = 2$, we see that we just need to determine what this map does in $H^2(C_3;\Bbb Z)$; it will either be the identity, or multiplication by $-1$; in the former case the induced map on $H^*(C_3;\Bbb Z)$ is the identity. In the latter case, $c^k \mapsto (-c)^k = (-1)^k c^k$, and so $i^*: H^{4n+2}(C_3;\Bbb Z)$ is multiplication by $-1$, and $i^*: H^{4n}(C_3;\Bbb Z)$ is the identity.
To pin this down, recall the universal coefficient theorem gives a natural isomorphism for any finite group $G$ $$H^2(G;\Bbb Z) \cong \text{Ext}(H_1(G;\Bbb Z), \Bbb Z),$$ because group cohomology in positive degrees is torsion, and in particular $$\text{Hom}(H_2(G;\Bbb Z), \Bbb Z) = 0.$$
Lastly, conclude with the fact that $H_1(G;\Bbb Z) \cong G^{\text{ab}}$ is a natural isomorphism, and $\text{Ext}(A, \Bbb Z)$ is naturally equivalent to $\text{Hom}(A, S^1)$ for any finite group $A$; essentially, this is saying that $\text{Ext}(A, \Bbb Z)$ is non-canonically isomorphic to $A$ itself, but that the induced maps are dualized; they go in the opposite direction.
In any case, applying all this to $G = C_3$, we find that the induced map of inversion on $H^2(C_3;\Bbb Z) = \text{Ext}(\Bbb Z/3, \Bbb Z)$ is multiplication by $-1$.
Now you need to know the group cohomology of $H^*(C_2;\Bbb Z/3)$, where $\Bbb Z/3$ is either given the trivial action or the negation action. Using the calculation here, as well as the fact that $H^0(G;M) = M^G$ by definition, we see that $H^*(C_2;\Bbb Z/3) = 0$ in positive degrees, and that $H^0(C_2;\Bbb Z/3) = \Bbb Z/3$ for the trivial action, but $H^0(C_2;\Bbb Z/3) = 0$ with the negation action.
(A high-tech calculation: if $G$ is a finite group and $M$ is a $G$-module in which multiplication by $|G|$ is an isomorphism, the existence of the transfer map implies that $H^*(G;M) = 0$ in positive degrees. In particular, this applies to $G = \Bbb Z/2$ and $M = \Bbb Z/3$ with any action.)
Now run your spectral sequence. The $q = 0$ line is $$\Bbb Z, 0, \Bbb Z/2, 0, \Bbb Z/2, 0, \cdots $$ while the $p = 0$ line is $$\Bbb Z, 0, 0, 0, \Bbb Z/3, 0, 0, 0, \Bbb Z/3, 0, \cdots $$
That is, there is a $\Bbb Z/3$ on every $E^{0,4n}_2$, and it is otherwise zero.
The spectral sequence is completely supported on these axes. Because everything is supported in even bidegree, there can be no nontrivial differentials, and $E_2 = E_\infty$. Along with the observation that there are no nontrivial extension problems (because $\Bbb Z/2$ and $\Bbb Z/3$ only give rise to a single abelian group as extension), we get the desired calculation.
If you also want to know what the product structure is, observe that the map $H^*(C_2;\Bbb Z) \to H^*(\Bbb S_3; \Bbb Z)$ sends the degree $2$ generator to the degree $2$ generator (you can see this at the level of the $E^2$ page of your spectral sequence), and is a ring homomorphism.
Similarly, one sees that the homomorphism $H^*(S_3; \Bbb Z) \to H^*(C_3; \Bbb Z)$ sends a 3-torsion generator in degree $4$ to the generator of $H^4(C_3;\Bbb Z)$.
These two facts, combined with the description of the underlying cohomology groups above, gives an isomorphism $$H^*(S_3;\Bbb Z) \cong \Bbb Z[c_1, c_2]/(2c_1, c_2-3c_1^2),$$ where $|c_i| = 2i$; this means that $c_1$ is $2$-torsion and $c_2 = 3c_1^2$ is $6$-torsion.
