Is the plane minus a line segment homeomorphic with punctured plane?

The map $(x,y)\mapsto (u,v)$, described below, is continuous on $\mathbb{R}^2\setminus \{(0,0)\}$. Indeed, each of the three pieces is continuous, and they agree on overlaps.
$$v=y,\qquad u=\begin{cases} x,\quad &x\le 0,\\ x/|y|,\quad & 0\le x\le |y|, \\ x-|y|+1,\quad &x>|y| \end{cases} $$ Also, the image of any point in $\mathbb{R}^2\setminus \{(0,0)\}$ is contained in $\mathbb{R}^2\setminus([0,1]\times \{0\})$.

The inverse of the aforementioned map is $$y=v,\qquad x=\begin{cases} u,\quad &x\le 0,\\ u|v|,\quad & 0\le u\le 1, \\ u+|v|-1,\quad &u>1 \end{cases} $$ It is continuous on all of $\mathbb{R}^2$; again, because the pieces agree on overlaps. The image of any point in $\mathbb{R}^2\setminus([0,1]\times \{0\})$ is contained in $\mathbb{R}^2\setminus \{(0,0)\}$.

The combination of stated properties implies the map is a desired homeomorphism.

Let $A=\mathbb R^2\setminus([0,1]\times\{0\})$ and $B=\mathbb R^2\setminus\{0,0\}.$ Place a coordinate system on $A$ by labeling each point $(\ell,\theta)$ where $\ell$ is the euclidean distance from the point to the line segment $L=[0,1]\times\{0\}$, and $\theta\in[0,2\pi)$ is the angle of the point from some fixed ray in $\mathbb R^2$. Put the standard polar coordinates on $B$. Define $f:A\to B$ by $f(\ell,\theta)=(\ell,\theta)$.

A simple convexity argument is enough to show bijectivity. Continuity in both directions is harder to show rigorously, but not hard to see.

For convenience, I will work with $\Bbb R^2\setminus([-1,1]\times\{0\})$ instead, which is obviously homeomorphic to your space by an affine transformation. The ellipse with foci at $\pm1$ and sum of distances to the foci equal to $2(r+1)$ has major axis $a=r+1$ and minor axis $b=\sqrt{r(r+2)}$, which yields the parameterization

$$x=(r+1)\cos\theta\\y=\sqrt{r(r+2)}\sin\theta$$

which is a homeomorphism from the $(r,\theta)$ polar parameterization of $\Bbb R^2\setminus\{(0,0)\}$ to the cartesian parameterization of $\Bbb R^2\setminus([-1,1]\times\{0\})$. (I find this parameterization preferable to those of @Soup and @AlexS. because it is not piecewise.)

From the picture, it looks like there is likely to be an underlying conformal map, if the $r$-parameterization is fixed to satisfy Cauchy-Riemann. Setting $r=f(\alpha)$ and equating the $\partial_\alpha$ and $\partial_\theta$ partial derivatives gives $f(\alpha)=2\sinh^2(\alpha/2)=\cosh\alpha-1$ as a conformal $(\alpha,\theta)$ parameterization, and plugging this back in gives the parameterization

$$(x,y)=(\cos\theta\cosh\alpha,\sin\theta\sinh\alpha)$$                                         

which is easily seen to be the conformal map corresponding to $f(z)=\cos z$. In other words, $\cos z$ is a homeomorphism from the quotient of the top half-plane $\{z\in\Bbb C\mid\Im z>0\}$ under the equivalence $z\sim w$ iff $z-w\in 2\pi\Bbb Z$, to $\Bbb C\setminus[-1,1]$; at the same time $f(x+iy)=ye^{ix}$ is a homeomorphism from this space to $\Bbb C\setminus\{0\}$. Note that the second map is not holomorphic, and indeed there is no such holomorphic map. They are not conformally equivalent because $\Bbb C\setminus\{0\}$ is the punctured plane and $\Bbb C\setminus[-1,1]$ is conformally equivalent to the punctured disk $\{z\mid0<|z|<1\}$ via the above $\cos z$ mapping and $e^{-iz}$, and the punctured plane and punctured disk are not conformally equivalent by the Riemann mapping theorem for doubly connected domains.

Without writing formulas this can be seen by noting that the universal cover in both cases is the disk, while the fundamental group in both cases is $\mathbb{Z}$. Therefore in the class of orientable manifolds, the quotient manifold must be the cylinder; i.e., both of your spaces are homeomorphic to an infinite cylinder.

This map should do: $$(x,y)\mapsto\begin{cases}(x+\operatorname{sgn}(x)(|y|-1),y),&|x|>1\\(xy,y),&|x|\le1\end{cases}$$