Let $X,Y,Z$ topological spaces. Suppose that product spaces $X \times Y$ and $X \times Z$ are homeomorphic. It's true that $Y$ and $Z$ are homeomorphics? Since that the projection map is only continuous I don't know if this result is really true.
I am interested in the validity of this result to conclude that $\mathbb{S}^2\times \mathbb{R}$ and $\mathbb{R}^3$ are not homeomorphic
$\mathbb{N} \times \mathbb{N}$ is homeomorphic to $\mathbb{N} \times \{1\}$ (all spaces considered with the discrete topology), because they are in bijection with each other and any map is continuous in the discrete topology (so any bijection will be a homeomorphism). Obviously $\{1\}$ is not homeomorphic to $\mathbb{N}$.
As for your second question, note that $(x_1, \cdots, x_{n+1}, z) \mapsto (e^{z} x_1, \cdots, e^z x_{n+1})$ is a homeomorphism between $\mathbb{S}^n \times \mathbb{R}$ and $\mathbb{R}^{n+1} \setminus \{0\}$, so an homeomorphism between $\mathbb{S}^n \times \mathbb{R}$ and $\mathbb{R}^{n+1}$ would imply an homeomorphism between $\mathbb{R}^{n+1}$ and $\mathbb{R}^{n+1} \setminus \{0\}$. But $\mathbb{R}^{n+1} \setminus \{0\}$ is not contractible (while $\mathbb{R}^n$ obviously is): indeed, it retracts onto $\mathbb{S}^n$, and $\mathbb{S}^n$ is not contractible. Therefore no such homeomorphism can exist, and in particular $\mathbb{S}^2 \times \mathbb{R}$ and $\mathbb{R}^3$ are not homeomorphic.