Considering the equivalence relation on the manifold $\mathbb{R}P^{2}\times \mathbb{R}$ which identifies all of $\{[1:0:0]\} \times \mathbb{R}$ and no other identifications, how can we give the quotient a structure of a differentiable manifold?
The motivation behind this construction is to view the quotient as copies of $\mathbb{R}P^{2}$ at each point (its origin) on the $x$-axis in $\mathbb{R}^{3}$, which corresponds to all lines in $\mathbb{R}^{3}$ that intersect the $x$-axis. Since $[1:0:0]$ in each copy describes the $x$-axis, we identify them. Since any other $[x_{1}:x_{2}:x_{3}]$ in one of the copies uniquely describes the line with direction $(x_{1},x_{2},x_{3})$ passing through a point on the $x$-axis, we do not identify these.
Ignoring the $x$-axis in this construction gives a manifold of dimension $3$, $(\mathbb{R}P^{2} \setminus \{[1:0:0]\})\times \mathbb{R}$, viewed as the set of all lines in $\mathbb{R}^{3}$ intersecting the $x$-axis not including the $x$-axis itself. I imagine that the aforementioned quotient is also of dimension $3$, but I'm struggling to come up with the appropriate charts.
The point lying under $[1:0:0]\times \Bbb{R}$ in your space has a connected neighbourhood $A$ homeomorphic to $\Bbb{R}^3/X$, where $X$ denotes the $x$-axis in $\Bbb{R}^3$. Removing the image of $A$ in $X$ gives a space with a non-trivial fundamental group. Removing a point from a $3$-manifold cannot affect the fundamental group of any open subset of the manifold. So your space cannot be a manifold.