Basis for $L^2(0,T;H)$ Given a basis $b_i$ for the separable Hilbert space $H$, what is the basis for $L^2(0,T;H)$? Could it be $\{a_jb_i : j, i \in \mathbb{N}\}$ where $a_j$ is the basis for $L^2(0,T)$?

You are not far from correct result. The desired basis is a family of fnctions 
$\{f_{i,j}:i,j\in\mathbb{N}\}$ defined as
$$
f_{i,j}(t)=a_j(t)b_j
$$
The deep reason for this is the following. Since we have an identification. 
$$
L_2((0,T), H)\cong L_2(0,T)\otimes_2 H
$$
it is enough to study to study bases of Hilbert tensor product of Hilbert spaces. It is known that for Hilbert spaces $K$, $H$ with orthnormal bases $\{e_i:i\in I\}$ and $\{f_j:j\in J\}$ respectively the family
$$
\{e_i\otimes_2 f_j:i\in I\; j\in J\}
$$
is an orthnormal basis of $K\otimes_2 H$
