The quotient space $Y = X / \sim$ as a set is just the set of equivalence classes of $X$ under $\sim$, so the set $\{ [x]: x \in \mathbb{R} \} $ in your case.
The equivalence class of a number $x$ is just (in your case) the set $\{ x+n : n \in \mathbb{Z} \}$. Now we need a topology. The standard topology that we take on $Y$ is all subsets $O$ of $Y$ (where points in $Y$ are "really" subsets of $X$, the equivalence classes) such that $q^{-1}[O]$ is open in the topology of $X = \mathbb{R}$. Here $q$ is the map that sends $x$ to its class $[x]$ in $Y$, the so-called quotient map. This is called the quotient topology on $Y$, and as you see it assumes you have a topology on $X$ already, and we give $Y$ the largest topology possible to still have $q$ continuous. (The smallest one would always be the indiscrete topology, which is not very interesting, hence the other "natural" choice.)
Now, if we have a function $f$ from $Y$, the quotient space in the quotient topology, to any space $Z$, then $f$ is continuous iff $f \circ q$ is continuous as a function from $X$ to $Z$: one way is clear, as the composition of continuous maps is continuous, and for the other side, if $O$ is open in $Z$, by definition $f^{-1}[O]$ is open in $Y$ iff $q^{-1}[f^{-1}[O]]$ is open in $X$, and this set equals $(f \circ q)^{-1}[O]$ which is open, as by assumption $f \circ q$ is continuous.
Now, consider the map $f$ that sends the class $[x]$ to the point $e^{2\pi ix}$ in $\mathbb{S}^1$, the unit circle.
This is well-defined: if $x'$ were another representative of $[x]$, then $x \sim x'$ and thus $x - x'$ is an integer and so $f(x') = f(x)$.
It is continuous, as $f \circ q$ is just the regular map sending $x$ to $e^{2\pi ix}$, and this is even differentiable etc.
It is clearly surjective and injective because the only way $[x]$ and $[y]$ will have the same value is when $2 \pi ix - 2 \pi i y$ is an integer multiple of $2 \pi i$, which happens iff $x - y$ is an integer.
One can also check that $q[X] = q[[0,1]]$ and by continuity of $q$ we have that $Y$ is compact. This makes (with $\mathbb{S}^1$ Hausdorff) the map $f$ a homeomorphism, by standard theorems.
We could also have achieved this as the quotient of $[0,1]$ under the equivalence relation that has exactly one non-trivial class, namely $\{0,1\}$. This is more intuitive, as we then glue together (consider as one point) just the points $0$ and $1$, and this geometrically gives a circle. In your example (which is a nice so-called covering map, and a group homomorphism as well) we glue a lot more points together, but all classes are now similar: just shifted versions of a point by an integer. We sort of wrap the interval $[0,1)$ infinitely many times over itself.
So you have
$$f:(X\times I)/\sim'\to (X/\sim)\times I$$
$$f([x,t])=([x],t)$$
and (as you've noted) this is a well defined continuous bijection. We can easily find the inverse:
$$g:(X/\sim)\times I\to (X\times I)/\sim'$$
$$g([x], t)=[x,t]$$
and the question is whether $g$ is continuous. So pick an open subset $U\subseteq (X\times I)/\sim'$ and let $\pi:X\times I\to (X\times I)/\sim'$ be the projection. Let $V:=\pi^{-1}(U)$. It is obviously open.
Now let $p:X\to X/\sim$ be the other projection and let $\tau:X\times I\to (X/\sim)\times I$ be given by $\tau(x,t)=(p(x), t)$. In other words $\tau=p\times id$ in your notation. Note that $\tau^{-1}(g^{-1}(U))=V$. Now if we knew that $\tau$ is a quotient map then we are done because that would imply that $g^{-1}(U)$ is open.
So when is $\tau$ a quotient map? For that we need $I$ to be locally compact and this is known as
Theorem (Whitehead): Let $X,Y,Z$ be topological spaces with $Z$ locally compact. If $f:X\to Y$ is a quotient map and $id:Z\to Z$ is the identity then $f\times id:X\times Z\to Y\times Z$ is a quotient map.
Best Answer
Not an answer, but a remark which is too important (in my opinion) to be just a comment: $\mathsf{Top}$ fails to satisfy several convenient categorical properties, such as the one discussed here that quotient maps don't have to be stable under products. Related to that, $\mathsf{Top}$ is not cartesian closed. However, in convenient categories of spaces all these properties hold (see here). An example is the category of compactly generated (weak) Hausdorff spaces $\mathsf{CGWH}$ (see here and there).
All this indicates that $\mathsf{Top}$ is the "wrong" category for doing topology. A better candidate is $\mathsf{CGWH}$. In practice, there is no additional effort when working in that "correct" category.
Notice that many books and papers in topology just use all these convenient properties although they claim to work with topological spaces. Most authors (seem to) ignore this problem, they even don't let the reader know what their category of "spaces" is (topological spaces? compact spaces? compactly generated spaces? CW complexes? simplicial sets? Almost everything has been called "spaces").