Solution:

Consider any two points $P_{1} = (x_{1}, y_{1})$ and $P_{2} = (x_{2}, y_{1})$ lying on the horizontal half-line given by the intersection of $y = y_{1}$ with the first quadrant (see also the figure, where $y_{1} = 1$). Let $r_{1}, r_{2}$ be the two half-lines starting at $P_{1}, P_{2}$ considered in the problem. These half-lines have the same slope with respect to the horizontal axis and are therefore parallel. This can also be seen in terms of Euclidean geometry, since the triangles formed by the points $P_{1}, (x_{1}+1, y_{1}), (x_{1}+1, 2y_{1})$ and $P_{2}, (x_{2}+1, y_{1}), (x_{2}+1, 2y_{1})$ are obtained from one another by a horizontal translation of $x_{2} - x_{1}$, and therefore their hypotenuses (which lie on the half-lines $r_{1}, r_{2}$) are parallel.

Up to symmetry, we now assume $x_{2} > x_{1}$, so that the half-line $r_{2}$ lies to the right of $r_{1}$. Choose any point $Q = (x_{Q}, y_{Q})$ on $r_{2}$ distinct from $P_{2}$. In particular $y_{Q} > y_{1}$, since $P_{2}$ is the point with smallest ordinate on the whole half-line $r_{2}$. The half-line $r$ starting at $Q$ and passing through $(x_{Q}+1, 2y_{Q})$ has slope $y_{Q}$, greater than the slope of the half-line $r_{1}$, and therefore $r_{1}$ and $r$ meet at some point $S$ to the right of $Q$.

It then follows from the hypotheses of the problem that $S$ has the same color as $Q$ (since $S$ belongs to the half-line $r$) and that $Q$ has the same color as $P_{2}$ (since it belongs to the half-line $r_{2}$). On the other hand, it is also true that $S$ belongs to the half-line $r_{1}$, and therefore has the same color as $P_{1}$. It follows therefore that $P_{1}, Q$ and $P_{2}$ all have the same color. Since the points $P_{1}, P_{2}$ were chosen arbitrarily on the line $y = y_{1}$, this shows that such a horizontal line consists entirely of points of the same color. Since $y_{1}$ was also chosen arbitrarily, we obtain that every horizontal half-line is monochromatic.

Finally, take any two horizontal half-lines $s_{1}, s_{2}$, given by the intersections with the first quadrant of the lines $y = y_{1}$ and $y = y_{2}$ respectively. Up to symmetry we may assume $y_{1} < y_{2}$. Consider any point $P$ on the half-line $s_{1}$. The half-line $r_{P}$ starting at $P$ with slope equal to the ordinate of $P$ intersects every horizontal line lying above $s_{1}$, and therefore in particular intersects $s_{2}$. Since the half-line $r_{P}$ is monochromatic and contains both a point of $s_{1}$ and a point of $s_{2}$, we obtain that the common color of all points of $s_{1}$ equals the common color of all points of $s_{2}$. Finally, this reasoning holds for any pair of horizontal half-lines, and therefore every point of the first quadrant is colored with the same color.

We conclude by including, for completeness, an algebraic verification of the (graphically evident) claim that the half-lines $r$ and $r_{1}$ actually meet at some point $S$. The lines containing the half-lines $r_{1}$, $r_{2}$ have respective equations $y = y_{1}(x - x_{1}) + y_{1}$, $y = y_{1}(x - x_{2}) + y_{1}$, where we assume $x_{2} > x_{1}$. The half-lines $r_{1}$ and $r_{2}$ are described by these equations, restricted however to values $x \geq x_{1}$ (respectively $x \geq x_{2}$).

If we call $(x_{Q}, y_{Q})$ the coordinates of the point $Q$ (with $x_{Q} > x_{2}$) we then have $y_{Q} = y_{1}(x_{Q} - x_{2}) + y_{1} = y_{1}(x_{Q} - x_{2} + 1)$, and the line containing the half-line $r$ has equation $y = y_{Q}(x - x_{Q} + 1)$ (the points of the half-line are those with $x \geq x_{Q}$). Putting this equation into a system with that of the line containing $r_{1}$ we find

$$\left\{ \begin{array}{l} y = y_{1}(x - x_{1} + 1) \\ y = y_{Q}(x - x_{Q} + 1) \end{array} \right.$$

from which we can obtain the $x$ coordinate of the intersection point, that is

$$\begin{aligned} x & = \frac{y_{1} - y_{Q}}{y_{Q} - y_{1}} + \frac{y_{Q} x_{Q} - x_{1} y_{1}}{y_{Q} - y_{1}} = -1 + \frac{y_{1}(x_{Q} - x_{2} + 1) x_{Q} - x_{1} y_{1}}{y_{1}(x_{Q} - x_{2})} \\ & = -1 + \frac{(x_{Q} - x_{2} + 1) x_{Q} - x_{1}}{x_{Q} - x_{2}} = -1 + \frac{x_{Q} - x_{1}}{x_{Q} - x_{2}} + x_{Q}. \end{aligned}$$

It suffices now to observe that $x_{1} < x_{2} < x_{Q}$ implies $x_{Q} - x_{1} > x_{Q} - x_{2} > 0$, so that the ratio $\frac{x_{Q} - x_{1}}{x_{Q} - x_{2}}$ is strictly greater than $1$, and therefore the $x$ coordinate of the intersection point computed above is strictly greater than $x_{Q}$ (hence also than $x_{1}$), that is, this point is indeed to the right of $Q$ and belongs to the half-lines $r$ and $r_{1}$.

Solution:

We show directly that any two points $P_{1} = (x_{1}, y_{1})$ and $P_{2} = (x_{2}, y_{2})$ of the first quadrant have the same color. Let $r_{1}, r_{2}$ be the half-lines starting at $P_{1}, P_{2}$ with respective slopes $y_{1}, y_{2}$.

We observe as a special case that if $r_{1}, r_{2}$ meet at a point $Q$, then clearly $P_{1}$ and $P_{2}$ have the same color, because $Q$ has the same color as $P_{1}$ (since it lies on $r_{1}$) and also the same color as $P_{2}$ (since it lies on $r_{2}$).

In the general case (that is, whether or not $r_{1}, r_{2}$ intersect) we can proceed as follows. Choose a point $S = (x_{S}, y_{S})$ satisfying the following conditions:
(a) $S$ is to the right of both $P_{1}$ and $P_{2}$;
(b) $S$ is below both $r_{1}$ and $r_{2}$;
(c) $y_{S}$ is greater than both $y_{1}$ and $y_{2}$.

We will show below, algebraically, that such a point always exists. The half-line $r_{S}$ starting at $S$ has slope $y_{S}$, greater than those of $r_{1}, r_{2}$, and since $S$ is below $r_{1}, r_{2}$ and to the right of $P_{1}, P_{2}$, the half-line $r_{S}$ intersects both $r_{1}$ and $r_{2}$, at two points which we call $S_{1}, S_{2}$ (that these half-lines actually intersect can be verified algebraically in a way very similar to what was done in the first solution). The common color of the points of $r_{1}$ is the color of $S_{1}$ and the common color of the points of $r_{2}$ is the color of $S_{2}$. But on the other hand $S_{1}, S_{2}$ belong to the same monochromatic half-line $r_{S}$, so $r_{1}, r_{2}$ are indeed of the same color. In particular, $P_{1}$ and $P_{2}$ are colored with the same color. Since this holds for every pair of points of the first quadrant, the entire first quadrant is colored with the same color.

To show the existence of the desired point $S = (x_{S}, y_{S})$, choose
$$x_{S} = (x_{1} + x_{2}) + \left(\frac{1}{y_{1}} + \frac{1}{y_{2}} + \frac{y_{2}}{y_{1}} + \frac{y_{1}}{y_{2}}\right)$$
and $y_{S} = \max\{y_{1}, y_{2}\} + 1$. Note that $x_{S}$ is greater than both $x_{1}$ and $x_{2}$, that is, $S$ lies to the right of $P_{1}$ and $P_{2}$. The lines containing the half-lines $r_{1}, r_{2}$ have respective equations
$$y = y_{1}(x - x_{1} + 1), \quad y = y_{2}(x - x_{2} + 1)$$
the half-lines $r_{1}, r_{2}$ are obtained by restricting to values $x \geq x_{1}$ (respectively $x \geq x_{2}$).

The points $T_{1}, T_{2}$ of $r_{1}, r_{2}$ with abscissa $x_{S}$ therefore have respective ordinates
$$y = y_{1}\left((x_{1} + x_{2}) + \left(\frac{1}{y_{1}} + \frac{1}{y_{2}} + \frac{y_{2}}{y_{1}} + \frac{y_{1}}{y_{2}}\right) - x_{1} + 1\right) > y_{1}\left(\frac{1 + y_{2}}{y_{1}} + 1\right) = y_{1} + y_{2} + 1$$
and
$$y = y_{2}\left((x_{1} + x_{2}) + \left(\frac{1}{y_{1}} + \frac{1}{y_{2}} + \frac{y_{2}}{y_{1}} + \frac{y_{1}}{y_{2}}\right) - x_{2} + 1\right) > y_{2}\left(\frac{1 + y_{1}}{y_{2}} + 1\right) = y_{1} + y_{2} + 1.$$

Note that since $y_{1}, y_{2}$ are positive we have $y_{1} + y_{2} + 1 > \max\{y_{1}, y_{2}\} + 1$. It follows that the point $S = (x_{S}, y_{S})$ has ordinate strictly smaller than $T_{1}, T_{2}$ and therefore actually lies below the half-lines $r_{1}, r_{2}$. On the other hand, by construction $S$ has ordinate strictly greater than $y_{1}, y_{2}$, so it has all the desired properties.