Let $z$ be the difference of two solutions. Then $z$ solves the homogeneous equation $z'' =pz' +qz $ with homogeneous boundary conditions.
If $z$ is not identically zero, then it's either strictly positive or strictly negative somewhere. Let's take on the first case, since we can switch to $-z$ anyway. At the point of absolute maximum (which is not an endpoint) we have $z' =0$, $z''\le 0$ and $qz>0$. Contradiction.
(Existence). I assume the existence for IVP is known. Let $y_0$ be the solution of the IVP $y_0(a)=A$, $y_0'(a)=0$ (and same DE as in the problem). Also, let $y_1$ be the solution of the homogeneous equation $y_1''=py_1'+qy$ with the initial conditions $y_1(a)=0$ and $y_1'(a)=1$. Let us observe that for any number $k$ the combination $y_0+ky_1$ solves your DE and satisfies the left-endpoint boundary condition.
It remains to choose $k$ so that $y_0(b)+ky_1(b)=B$. Clearly, this is possible as long as $y_1(b)\ne 0$. And how do we know that $y_1(b)\ne 0$? Because if $y_1(b)=0$, then $y_1$ and $\equiv 0$ both solve the homogeneous BVP for $y''=py'+qy$, contradicting uniqueness.