Suppose as an example I have the magnitude of an electric $\left \| \vec{E} \right \|=\frac{\lambda}{2\epsilon s\pi}$ This is the equivalent to $\vec{E}=\frac{\lambda}{2\epsilon s\pi}\hat{r}$.
Is this correct? But why? I haven't come across this signalling before.
Exploit the symmetry of the problem, which you already have done when applying Gauss's Law (not Gaussian's Law).
In your development, you assumed that $\vec E$ has only a radial component $\hat \rho$ and depends only on the radial variable $\rho$.
Thus, we can write the electric field as $$\vec E=\hat \rho E_{\rho}(\rho)\tag 1$$
Then, you constructed a cylindrical surface with height $L$, centered on the $z$ axis, encompassing part of the line charge.
The total charge $Q$ enclosed is $Q=\lambda \, L$.
Finally, Gauss's Law states $$\oint_S \vec E\cdot \hat n \,dS=Q/\epsilon_0\tag 2$$
Using $(1)$ in $(2)$ reveals
$$\int_0^L\int_0^{2\pi}\hat \rho E_{\rho}(\rho)\cdot \rho \rho d\phi\,dz=\lambda\,L/\epsilon_0\implies E_{\rho}(\rho)=\frac{\lambda }{2\pi\epsilon_0\,\rho} \tag 3$$
Finally, substituting $(3)$ into $(1)$ yields
$$\vec E=\hat \rho \frac{\lambda }{2\pi\epsilon_0\,\rho}$$