Does $X\times S^1\cong Y\times S^1$ imply that $X\times\mathbb R\cong Y\times\mathbb R$?
No, it is not possible to conclude that $X\times\mathbb{R}\cong Y\times\mathbb{R}$ from $X\times S^1\cong Y\times S^1$, even in the case where $X$ and $Y$ are compact differentiable manifolds. In fact, Charlap1 showed the following in 1965.
There exist compact manifolds $X$, $Y$ of different homotopy types, but with $X\times S^1$ diffeomorphic to $Y\times S^1$.
As $X\times\mathbb{R}$ has the same homotopy type as $X$ for any topological space $X$, this implies that $X\times\mathbb{R}$ and $Y\times\mathbb{R}$ have different homotopy types, so they are certainly not homeomorphic. See also Hilton, Mislin & Roitberg2 for examples of non-homotopy equivalent manifolds with diffeomorphic products with higher dimensional spheres.
I just found those references by a bit of searching around, so I am not familiar with all of the details. However, I will now construct an explicit example of compact manifolds with different fundamental groups (hence, different homotopy types), but with diffeomorphic products with the circle. I think this example works along similar lines to the examples which could be constructed with the results of Charlap. In my example, $X$ and $Y$ will be 10-dimensional manifolds with covering space $S^9\times\mathbb{R}$.
Since the products $X\times S^1$ and $Y\times S^1$ are to be diffeomorphic, they must have the same fundamental group. So, the first thing is to find two non-isomorphic groups $G$, $H$ whose direct products with the infinite cyclic group $Z$ (i.e., the integers under addition) are isomorphic. As it is always possible to cancel direct products with $Z$ for abelian groups, it is necessary that $G$ and $H$ are non-abelian. Expressed in terms of generators and relations, one such example is, $$ \begin{align} G &= \langle x,y\mid yx = xy^2, y^{32}=y\rangle,\cr H &= \langle x,y\mid yx = xy^4, y^{32}=y\rangle. \end{align} $$ This is a very slightly simplified version of the example given in Hirshon3. Note that each element of $G$ (resp., $H$) can be uniquely expressed as $x^ry^s$ with $s$ taken modulo 31, and satisfy the multiplication rule $(x^{r_1}y^{s_1})(x^{r_2}y^{s_2})=x^{r_1+r_2}y^{2^{r_2}s_1+s_2}$ (resp., $x^{r_1+r_2}y^{4^{r_2}s_1+s_2}$). These groups are not isomorphic. Indeed, if $\bar x$, $\bar y$ are elements generating $G$ with $\bar y$ of order 31, then we must have $\bar y$ a power of $y$ and $\bar x=x^{\pm1}y^s$, so that $\bar x^{-1}\bar y\bar x = \bar y^2$ or $\bar y^{16}$. In either case, this does not equal $\bar y^4$, so $G\not\cong H$.
On the other hand, the direct product $G\times Z$ can be formed by adding an additional generator $z$ to $G$ which commutes with $x$ and $y$ and satisfies no further relations. Setting $\bar x = x^2z$ and $\bar z = x^5z^2$, then $\bar x, y, \bar z$ also generate $G\times Z$, $\bar z$ commutes with $\bar x$ and $y$, and $\bar x,y$ satisfy the relation $y\bar x=\bar xy^4$. From this it can be seen that $G\times Z\cong H\times Z$.
I'll now define the spaces $X,Y$. The simply connected manifold $\hat X\equiv S^9\times\mathbb{R}$ can be realized as the submanifold of $\mathbb{C}^5\times\mathbb{R}$ consisting of the points $(z,s)$ with $\lVert z\rVert = 1$. Define the diffeomorphisms $R,S$ on $\hat X$ as $$ \begin{align} & R(z_0,z_1,z_2,z_3,z_4,s)=(\omega z_0,\omega^2z_1,\omega^4z_2,\omega^8z_3,\omega^{16}z_4,s),\cr & S(z_0,z_1,z_2,z_3,z_4,s)=(z_4,z_0,z_1,z_2,z_3,s+1) \end{align} $$ where $\omega=e^{2\pi i/31}$. These satisfy the relations $RS=SR^2$ and $R^{32}=R$, so the groups $\Lambda_G\equiv\langle S,R\rangle$ and $\Lambda_H\equiv\langle S^2,R\rangle$ are isomorphic to $G$ and $H$ respectively. Define $X$ and $Y$ to be the quotients $\hat X/\Lambda_G$ and $\hat X/\Lambda_H$. These are compact manifolds with non-isomorphic fundamental groups $\Lambda_G\cong G$ and $\Lambda_H\cong H$.
Finally, I'll show that $X\times S^1$ and $Y\times S^1$ are diffeomorphic. Note that $X\times S^1$ is just the manifold $X\times \mathbb{R}$ quotiented out by the translations $(x,t)\mapsto(x,t+n)$ ($n\in\mathbb{Z}$). This can be written as a quotient $\hat X\times\mathbb{R}/\Lambda^\prime$, where $\Lambda^\prime=\langle S\times I, R\times I, \hat I\times T\rangle$. Here, $\hat I, I$ are the identities on $\hat X$ and $\mathbb{R}$, and $T$ is the translation on $\mathbb{R}$ given by $t\mapsto t+1$. However, writing $\bar S=S^2\times T$ and $\bar T=S^5\times T^2$, then $\bar S, R\times I,\bar T$ generate $\Lambda^\prime$ and we have $$ \begin{align} U^{-1}\bar SU&=S^2\times I,\cr U^{-1}(R\times I)U&=R\times I,\cr U^{-1}\bar TU&=\hat I\times T, \end{align} $$ where $U$ is the diffeomorphism on $S^9\times\mathbb{R}\times\mathbb{R}$ given by $U(z,s,t)=(z,s+5t,s/2+2t)$. So, with $\cong$ denoting diffeomorphism, $$ \begin{align} X\times S^1 &\cong S^9\times\mathbb{R}\times\mathbb{R}/\langle S\times I,R\times I,\hat I\times T\rangle\cr &\cong S^9\times\mathbb{R}\times\mathbb{R}/\langle S^2\times I,R\times I,\hat I\times T\rangle\cr &\cong Y\times S^1. \end{align} $$
Aside: As was noted in the question and shown in the linked lecture series, the implication $X\times\mathbb{R}\cong Y\times\mathbb{R}\Rightarrow X\times S^1\cong Y\times S^1$ holds for compact spaces $X,Y$. However, it does not hold if the spaces are not compact. Although there are rather involved counterexamples, such as $X=\mathbb{R}^3$ and $Y$ being the Whitehead manifold, there are also much simpler counterexamples, and I'll just mention one here that I thought of. Take $X$ to be the sphere minus three points and $Y$ to be a torus minus one point. Then, $X$ is just the same as the open disc minus two points, so $X\times S^1$ embeds in $\mathbb{R}^3$. On the other hand, there exists an embedded closed surface and curve in $Y\times S^1$ with intersection number 1 (if it was embeddable in $\mathbb{R}^3$, the intersection number would have to be zero). So, $X\times S^1\not\cong Y\times S^1$. I'll leave the construction of the closed surface and curve as an interesting exercise, and also the homeomorphism showing that $X\times\mathbb{R}\cong Y\times\mathbb{R}$.
1Compact Flat Riemannian Manifolds: I, Leonard S. Charlap, Annals of Mathematics Second Series, Vol. 81, No. 1 (Jan., 1965), pp. 15-30. (link)
2Sphere Bundles Over Spheres and Non-Cancellation Phenomena, P. Hilton, G. Mislin & J. Roitberg. J. London Math. Soc. (1972) s2-6(1): 15-23 (link)
3On Cancellation in Groups, R. Hirshon, The American Mathematical Monthly. Vol. 76, No. 9 (Nov., 1969), pp. 1037-1039. (link) (Alt. freely available link from R. Hirshon's home page)