I think your best bet is to solve for $p_C$, and compute $1 - p_C$. I know that this isn't what you're looking for particularly, but practically speaking, this will be the more efficient approach.
As I understand your question, you seem to want to know if there are quick manipulations, possibly involving matrix multiplication, where we could extract the information about $p_A + p_B$. I don't think what you're looking for is feasible.
If we include the $p_A + p_B + p_C = 1$ restriction, we would get the system of linear equations:
$$\begin{bmatrix} 1 & 1 & 1 \\ a+ m & -b & -s\\ -a & b+ d & 0\\ -m & -d & s \end{bmatrix}\begin{bmatrix} p_{A}\\ p_{B}\\ p_{C}\\ \end{bmatrix}= \begin{bmatrix} 1\\ 0\\ 0\\ 0 \end{bmatrix}.$$
Note: the final row is just the negative of the sum of the second and third rows, so it is redundant. We could simply write:
$$\begin{bmatrix} 1 & 1 & 1 \\ a+ m & -b & -s\\ -a & b+ d & 0 \end{bmatrix}\begin{bmatrix} p_{A}\\ p_{B}\\ p_{C}\\ \end{bmatrix}= \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}.$$
Now, in order to suitably extract $p_A + p_B$, we would want a change of variables, which amounts to a change of basis. Let's call our new variables $x, y, z$. We want one of our new variables, $z$ say, to be $p_A + p_B$. We're hoping that, in transforming the system, we can simply read off the value of $z$, so that we can find $z$ without having to perform a full solving procedure. As I said, I don't think this is feasible.
To illustrate this, let's choose a change of variables in this vein: $x = p_B$, $y = p_C$, and $z = p_A + p_B$. Then $p_A = z - x$, $p_B = x$, and $p_C = y$, or in other words,
$$\begin{bmatrix}
p_A \\ p_B \\ p_C
\end{bmatrix} =
\begin{bmatrix}
-1 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0
\end{bmatrix}
\begin{bmatrix}
x \\ y \\ z
\end{bmatrix}.$$
Our system would become:
$$\begin{bmatrix} 1 & 1 & 1 \\ a+ m & -b & -s\\ -a & b+ d & 0 \end{bmatrix}
\begin{bmatrix}
-1 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0
\end{bmatrix}
\begin{bmatrix} x\\ y\\ z\\ \end{bmatrix}= \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix},$$
i.e.
$$\begin{bmatrix} 0 & 1 & 1 \\ -a - m - b & -s & a + m\\ a + b + d & 0 & -a \end{bmatrix}
\begin{bmatrix} x\\ y\\ z\\ \end{bmatrix}= \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}.$$
The value of $z$ is not easy to read from the above system without going down the path of Gaussian elimination. But maybe this was down to a bad choice of variable change? If we had chosen $x$ and $y$ more cleverly, could we have read off the value of $z = p_A + p_C$ more readily?
No, this is not really possible. In order to read off the value of $z$ easily, we would need one of the rows of the matrix to be of the form $[0, 0, \alpha]$ for some $\alpha \neq 0$, giving us an equation of the form $\alpha z = ?$, where $?$ is $0$ or $1$ taken from $\begin{bmatrix}1\\0\\0\end{bmatrix}$, depending if the row is the first row, or the second/third row. If the $?$ happens to be $0$, then this could only be possible if $p_A + p_B = 0$, which seems unlikely (especially since both are presumably probabilities). There may be some clever choices of $x$ and $y$ that would make the top row into $[0, 0, \alpha]$, but they would involve specific formulas in terms of the unknowns in the matrix, i.e. $a, b, d, m, s$, and finding such clever choices would be a problem at least as hard as solving the original system!
If you think, perhaps, that a little manipulation of the $\begin{bmatrix}1\\0\\0\end{bmatrix}$ is necessary, i.e. by multiplying by matrices on the left, then you're right, but this is exactly what Gaussian elimination does.
All of this to say, I think your idea is a dead end.
