The most accurate way to correct your data, I believe, would be to rotate it around the intersection of $y=x$ and $y=0.3x+0.5$ until it lines up with $y=x$.
Firstly, note that a line passing through the origin with slope $m$ makes an angle of $\arctan(m)$ with the $x$-axis. Thus, the angle between the two lines on your graph is $$\arctan(1)-\arctan\left(\frac{3}{10}\right)$$
Next, to find the coordinates $(x',y')$ of any point $(x,y)$ rotated by an angle $\phi$ around the origin, you can perform the following matrix multiplication:
$$\begin{bmatrix}
x' \\
y'
\end{bmatrix}
=
\begin{bmatrix}
\cos(\phi) & -\sin(\phi) \\
\sin(\phi) & \cos(\phi)
\end{bmatrix}
\begin{bmatrix}
x \\
y
\end{bmatrix}$$
Unfortunately, we're not rotating around the origin. That makes things a little more complicated. There's no way to represent that with $2\times2$ matrices, so we need to use $3\times3$ matrices. To find the coordinates $(x',y')$ of any point $(x,y)$ rotated by an angle $\phi$ around the point $(x_0,y_0)$ (lots of different $x$'s and $y$'s there; my apologies), you can perform the following matrix multiplication, which is a little scarier-looking but is much less scary than it looks:
$$\begin{bmatrix}
x' \\
y' \\
1
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 & x_0 \\
0 & 1 & y_0 \\
0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
\cos(\phi) & -\sin(\phi) & 0 \\
\sin(\phi) & \cos(\phi) & 0 \\
0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
1 & 0 & -x_0 \\
0 & 1 & -y_0 \\
0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
x \\
y \\
1
\end{bmatrix}$$
What's really going on here is that the first matrix to act on the original point (which is the right-most matrix) performs a coordinate transformation that moves the origin to the point around which we want to rotate, the middle matrix performs the rotation around that point, and then the left-most matrix undoes the coordinate transformation.
Multiplying the three matrices together, we find that the above equation simplifies a bit to:
$$\begin{bmatrix}
x' \\
y' \\
1
\end{bmatrix}
=
\begin{bmatrix}
\cos(\phi) & -\sin(\phi) & x_0(1-\cos(\phi))+y_0\sin(\phi)\\
\sin(\phi) & \cos(\phi) & y_0(1-\cos(\phi))-x_0\sin(\phi)\\
0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
x \\
y \\
1
\end{bmatrix}$$
Performing the final multiplication, simplifying a bit, and getting rid of the now-unnecessary third row, we find that
$$\begin{bmatrix}
x' \\
y'
\end{bmatrix}
=
\begin{bmatrix}
(x-x_0)\cos(\phi)-(y-y_0)\sin(\phi)+x_0 \\
(x-x_0)\sin(\phi)+(y-y_0)\cos(\phi)+y_0
\end{bmatrix}\tag{$\large\star$}$$
Theoretically speaking, this isn't a one-step process, but it can be sped up greatly by writing a simple program to do this or by spending a bit of time with a calculator. For each point on your line, you just need to plug the $x$- and $y$-coordinates of the point into the above equation to find the coordinates of the transformed point (which should lie on or near the line $y=x$).
The two lines intersect at $\left(\frac{5}{7},\frac{5}{7}\right)$, so $$x_0=y_0=\frac{5}{7},$$ and $$\phi=\arctan\left(\frac{3}{10}\right)-\arctan(1).$$ I know $\phi$ is reversed from what I said above, but just trust me that it needs to be this way for the math to work out. It's the same angle, just in a different direction.
I've done most of the calculations in case you just want a single matrix by which to multiply every point; all you need to do is plug in an $x$ and a $y$ and multiply (unfortunately, the third row is still necessary to make the multiplication possible, even though it's all $0$'s and $1$'s).
$$\begin{bmatrix}
x' \\
y' \\
1
\end{bmatrix}
=
\frac{1}{\sqrt{218}}
\begin{bmatrix}
13 & -7 & \frac{5}{7}\Big(\sqrt{218}-6\Big) \\
7 & 13 & \frac{5}{7}\Big(\sqrt{218}-20\Big) \\
0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
x \\
y \\
1
\end{bmatrix}$$
Alternatively:
If you want to correct the trend line instead of correcting every point individually, you can replace $y$ in the equation $y=0.3x+0.5$ with $$(x-x_0)\sin(\phi)+(y-y_0)\cos(\phi)+y_0$$ and replace $x$ with $$(x-x_0)\cos(\phi)-(y-y_0)\sin(\phi)+x_0$$ to end up with the equation
$$(x-x_0)\sin(\phi)+(y-y_0)\cos(\phi)+y_0=0.3\Big((x-x_0)\cos(\phi)-(y-y_0)\sin(\phi)+x_0\Big)+0.5$$
Those expressions should look familiar from equation $\left(\large{\star}\right)$. Once you plug in the appropriate values for $x_0$, $y_0$, and $\phi$ and then simplify, you should end up with the line $y=x$.
