Over spheres, you actually have a nice way to distinguish bundles.
Every sphere $S^n$ can be written as a union of two contractible spaces, $U \cup V$, whose intersection is a small neighborhood of the equator, $\mathbb{R} \times S^{n-1}$. For instance if $x,y$ are the north and south poles, respectively, you can take $U = S^n - \{x\}, V = X^n - \{y\}$.
Over $U$ and $V$, any bundle is trivial, as you noted. The bundle on the sphere is glued together from these two trivial bundles.
If you're interested in a vector bundle, the transition function for the bundle is given by a map $U \cap V \to GL_m$, for a vector bundle of rank $m$. If this map is null-homotopic, meaning that it can be homotoped to the constant map, then your bundle is trivial, as any trivialization over $U$ can then be glued to a trivialization over $V$. So in the end you are reduced to the study of maps from $U \cap V \simeq S^{n-1}$ to your structure group -- i.e., $\pi_{n-1}GL_m$.
For $m=1$, this amounts to asking whether any map $S^{n-1} \to GL_1$ is null-homotopic. In your case for $S^3$, since $GL_1(\mathbb{R})$ is a disjoint union of two contractible spaces, and $S^2$ is connected, any map is null-homotopic.
In the fiber bundle case with fiber $S^1$ and base $S^3$, you are looking at maps $S^2 \to S^1$. But $\pi_2(S^1) = *$ so any circle bundle is contractible over $S^3$.
Note that both facts hold for higher spheres as well.
You have defined $\Psi:B\times \mathbb R\to E$ by $(b,\lambda)\mapsto\lambda \sigma(b).$ Take a $b\in U\subseteq B$ and a local trivialization $\Phi:\pi^{-1}(U)\to U\times \mathbb R.$ Then, $\Phi\circ \sigma|_U:U\to U\times \mathbb R$ is continuous so $\Phi\circ \sigma|_U(b)=(b,f(b))$ for some continuous $f:U\to \mathbb R$ and so $\Phi\circ \Psi(b,\lambda)=\Phi(\lambda\sigma (b))=(\lambda\sigma(b),f(\lambda\sigma (b)).$ This means that $\Phi\circ \Psi$ is continuous, and therefore, since $\Phi$ is a homeomorphism, that $\Psi$ is continuous, too.
Best Answer
First of all, $\pi^{-1}(b) \neq s(b)$; $\pi^{-1}(b) \cong \mathbb{R}$ while $s(b) \in \mathbb{R}$.
You have defined a vector bundle homomorphism $F : B\times\mathbb{R} \to E$. Now you just have to show that it is in fact an isomorphism, i.e. that the map $F_b : \{b\}\times\mathbb{R} \to E_{b}$ is an isomorphism of vector spaces. The map $F_b$ is given by $(b, \lambda) \mapsto \lambda s(b)$. Is this linear map injective? That is, if $\lambda s(b) = 0$, is $\lambda$ necessarily zero?