Projection onto a line that doesn’t pass through the origin is not a linear transformation, but it is an affine transformation. This means that it can be represented by a matrix, but you need to use a $3\times3$ matrix and homogeneous coordinates.
There are several ways to build this matrix. Since your approach of computing the projections of the basis vectors does work for a line through the origin, let’s take advantage of that method by adding a couple of translations: first translate so that the line passes through the origin, project onto the translated line, then translate back. Any point on the line will do for these translations, so we’ll use the $y$-intercept $\mathbf t=(0,-3)$ since it can be read directly from the equation of the line. Letting $P_2$ stand for the $2\times2$ matrix of the projection onto the translated line, the projection onto the original line is then (in block form): $$P=\left[\begin{array}{c|c}I_2 & \mathbf t \\ \hline \mathbf 0^T&1 \end{array}\right]\left[\begin{array}{c|c}P_2&0\\\hline\mathbf 0^T&1\end{array}\right]\left[\begin{array}{c|c}I_2&-\mathbf t \\\hline \mathbf 0^T&1\end{array}\right] = \left[\begin{array}{c|c} P_2 & \mathbf t-P_2\mathbf t \\\hline \mathbf 0^T&1\end{array}\right].$$ Here, $I_2$ stands for the $2\times2$ identity matrix. The columns of $P_2$ are the projections of $(1,0)$ and $(0,1)$, respectively, onto the translated line, giving $$P_2=\left[\begin{array}{rr}\frac15&-\frac25\\-\frac25&\frac45\end{array}\right]$$ and $\mathbf t-P_2\mathbf t=\left(-\frac65,-\frac35\right)$ (I’ll leave it to you to work out for yourself and verify these values), so the full projection matrix is $$P=\left[\begin{array}{rrr}\frac15&-\frac25&-\frac65 \\ -\frac25&\frac45&-\frac35 \\ 0&0&1\end{array}\right].$$ To use this matrix, you need to append a $1$ to the coordinates of the point being transformed and then drop that $1$ after you’ve found the product, e.g., $$\left[\begin{array}{rrr}\frac15&-\frac25&-\frac65 \\ -\frac25&\frac45&-\frac35 \\ 0&0&1\end{array}\right]\begin{bmatrix}8\\3\\1\end{bmatrix}=\begin{bmatrix}-\frac45\\-\frac75\\1\end{bmatrix}.$$ Since the last component of the product $P(x,y,1)^T$ is always going to be $1$, you can drop the last row of the matrix instead if you like. You can also verify that the projections of $(1,0)$ and $(0,1)$ computed with this matrix are the same as the values that you were getting “by hand.”
Observe that $\mathbf t-P_2\mathbf t$ is the orthogonal rejection of $\mathbf t$ from the translated line. In other words, it’s the component of $\mathbf t$ perpendicular to the line, but this is just the perpendicular displacement of the original line from the origin. (That’s why we could take any point on the line for $\mathbf t$). This shift perpendicular to the line is what was missing from your initial attempt.
You can easily check for A considering the product by the basis vector of the plane, since $\forall v$ in the plane must be:
$$Av=v$$
Whereas for the normal vector:
$$An=0$$
Note that with respect to the basis $\mathcal{B}:{c_1,c_2,n}$ the projection matrix is simply:
$$P_{\mathcal{B}}=\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 0\end{bmatrix}$$
If you need the projection matrix with respect to another basis you simply have to apply a change of basis to obtain the new matrix.
For example with respect to the canonical basis, lets consider the matrix M which have vectors of the basis $\mathcal{B}:{c_1,c_2,n}$ as colums:
$$M=\begin{bmatrix}
-1 & 0 & 1\\
0 & -1 & 1\\
1 & 1 & 1\end{bmatrix}$$
If w is a vector in the basis $\mathcal{B}$ its expression in the canonical basis is $v$ give by:
$$v=Mw\implies w=M^{-1}v$$
Thus if the projection $w_p$ of w in the basis $\mathcal{B}$ is given by:
$$w_p=P_{\mathcal{B}}w$$
The projection in the canonical basis is given by:
$$M^{-1}v_p=P_{\mathcal{B}}M^{-1}v\implies v_p=MP_{\mathcal{B}}M^{-1}v $$
Thus the matrix:
$$A=MP_{\mathcal{B}}M^{-1}=$$
$$=\begin{bmatrix}
-1 & 0 & 1\\
0 & -1 & 1\\
1 & 1 & 1\end{bmatrix}\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 0\end{bmatrix}\begin{bmatrix}
-1 & \frac13 & \frac13\\
\frac13 & -1 & \frac13\\
\frac13 & \frac13 & \frac13\end{bmatrix}=\begin{bmatrix}
2/3 & -1/3 & -1/3\\
-1/3 & 2/3 & -1/3\\
-1/3 & -1/3 & 2/3\end{bmatrix}$$
represent the projection matrix in the plane with respect to the canonical basis.
Suppose now we want find the projection matrix from the base $\mathcal{B}$ to the canonical $\mathcal{C}$.
Let's consider the projection $w_p$ of w in the basis $\mathcal{B}$ is given by:
$$w_p=P_{\mathcal{B}}w$$
thus:
$$M^{-1}v_p=P_{\mathcal{B}}w\implies v_p=MP_{\mathcal{B}}w$$
Thus the matrix:
$$C=MP_{\mathcal{B}}=$$
$$=\begin{bmatrix}
-1 & 0 & 1\\
0 & -1 & 1\\
1 & 1 & 1\end{bmatrix}\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 0\end{bmatrix}=\begin{bmatrix}
-1 & 0 & 0\\
0 & -1 & 0\\
1 & 1 & 0\end{bmatrix}$$
represent the projection matrix from the base $\mathcal{B}$ to the canonical $\mathcal{C}$.
Best Answer
Your answer is correct...the diagonal form of a projection matrix always has only 1's and $0$'s on the diagonal...if you think about it this it makes sense, since vectors in the projection space is perfectly preserved, and vectors orthogonal to it will vanish.
To get the matrix representation of $T$ with respect to the canonical basis you just use the familiar similarity relation to find a diagonal representation, and since you already have this diagonal representation and the ordered basis that "creates" it, you have \begin{equation} [T]_\epsilon=\begin{bmatrix} 1 & -2\\ 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0\\ 0 & 0 \end{bmatrix}\begin{bmatrix} 1 & -2\\ 2 & 1 \end{bmatrix}^{-1}, \end{equation} where $\epsilon$ is the canonical basis