I've been trying to prove that $\text{Isom}(\mathbb{S}^2 \times \mathbb{R}) \cong \text{Isom}(\mathbb{S}^2) \times \text{Isom}(\mathbb{R})$ and I think I've come up with a simple proof, but it's so simple that I'm almost certain I made a mistake somewhere. Here goes:
Let $F: \mathbb{S}^2 \times \mathbb{R} \to \mathbb{S}^2 \times \mathbb{R}$ be an isometry. For each $(\hat{x}, y) \in \mathbb{S}^2 \times \mathbb{R}$, with $\hat{x} = (x_1, x_2, x_3) \in \mathbb{S}^2$ and $y \in \mathbb{R}$, writing $F(\hat{x}, y) = (A(\hat{x}), B(y))$, we want to prove that $A: \mathbb{S}^2 \to \mathbb{S}^2$ and $B: \mathbb{R} \to \mathbb{R}$ are isometries, that is, for any $s, t \in \mathbb{S}^2$ we have $\langle A(s) - A(t), A(s)- A(t) \rangle = \langle s - t, s -t \rangle$ and for all $n, m \in \mathbb{R}$ we have $|B(n) - B(m)| = |n - m|$.
Well, since $F$ is an isometry of $\mathbb{S}^2 \times \mathbb{R}$, given $s, t \in \mathbb{S}^2$, it must be true that $\langle F(s, 0) - F(t, 0),F(s, 0) - F(t, 0) \rangle = \langle(s - t, 0), (s-t, 0) \rangle$. It follows that:
$$\|A(s) - A(t)\|^2 + (B(0) - B(0))^2 = \|s-t\|^2 + (0-0)^2$$
and therefore $\langle A(s) - A(t), A(s)- A(t) \rangle = \langle s - t, s -t \rangle$, as we wanted to prove. The proof that $B$ is an isometry is analagous.
What's the catch here?
I will give a sketch that every isometry $F:S^2\times \mathbb{R}\to S^2\times \mathbb{R}$ is of the form $F(p,y)=(A(p),B(y)).$
Fix arbitrary $p\in S^2$ and $y\in \mathbb{R}$. Let $q\in S^2$ and $r\in \mathbb{R}$ be given by $(q,r)=F(p,y)$. Observe that the tangent space at $(p,y)$ splits as an orthogonal sum \begin{equation*} T_{(p,y)}(S^2\times\mathbb{R})=T_pS^2\oplus T_y\mathbb{R} \end{equation*} and similarly \begin{equation*} T_{(q,r)}(S^2\times\mathbb{R})=T_qS^2\oplus T_r\mathbb{R}. \end{equation*} Recall that $(dF)_{(p,y)}:T_pS^2\oplus T_y\mathbb{R}\to T_qS^2\oplus T_r\mathbb{R}$ is an isometry between vector spaces. Observe that \begin{equation*} (dF)_{(p,y)}\big(T_pS^2\big)\subset T_qS^2, \end{equation*} since otherwise $F$ would map some periodic geodesics to non-periodic geodesics. Since $T_pS^2$ and $T_qS^2$ have the same dimension, we even get \begin{equation*} (dF)_{(p,y)}\big(T_pS^2\big)=T_qS^2 \end{equation*} by the injectivity of $(dF)_{(p,y)}$. As $(dF)_{(p,y)}$ is an isometry, \begin{equation*} (dF)_{(p,y)}\big(T_y\mathbb{R}\big)=T_r\mathbb{R} \end{equation*} follows from the orthogonal decompositions of the tangent spaces. Hence we get a decomposition \begin{equation*} (dF)_{(p,y)}=a\oplus b \end{equation*} for some isometries $a:T_pS^2\to T_qS^2$ and $b: T_y\mathbb{R}\to T_r\mathbb{R}$. Therefore, for any $v\in T_pM$ and $c,t\in \mathbb{R}$ we have \begin{equation*} F\big(\exp_p(tv),y+ct\big)=\Big(\exp_q\big(ta(v)\big),r+tb(c)\Big). \end{equation*} Can you conclude from here?