The answer is no. Let $\kappa$ be any infinite cardinal, regular or singular, and assume for a contradiction that there is a set $E\subseteq\mathcal P(\kappa)$ satisfying your conditions. I will call the elements of $\kappa$ points and the elements of $E$ lines.
The lines do not all go through one point: Given a point $\alpha$, choose a point $\beta\ne\alpha$ and a point $\gamma$ not on the line through $\alpha$ and $\beta$; the line through $\beta$ and $\gamma$ does not go through $\alpha$.
There are $\lt\kappa$ lines through any point: Consider any point $\alpha$ and let $\lambda$ be the number of lines through $\alpha$. Choose a line $e$ which does not go through $\alpha$. Since each line through $\alpha$ meets $e$ in a different point, $\lambda\le|e|\lt\kappa$.
Now choose two distinct points $\alpha$ and $\beta$. Say there are $\lambda$ lines through $\alpha$ and $\mu$ lines through $\beta$. Let $e$ be the line through $\alpha$ and $\beta$. Now every point which is not on the line $e$ is the point of intersection of a line through $\alpha$ and a line through $\beta$. Hence $\kappa\le|e|+\lambda\mu\lt\kappa$ which is absurd.
This argument is adapted from the proof that a finite projective plane of order $n$ has $n^2+n+1$ points. In that case we have $|e|=\lambda=\mu=n+1$ and the number of points is exactly $|e|+(\lambda-1)(\mu-1)=n^2+n+1$.
P.S. More generally, the answer is still no if condition (2) is replaced by "for every $\alpha\in\kappa$ we have $|\{e\in E:\alpha\in e\}|\gt1$"; that was proved in my answer to this old question.