I'm assuming the matrices are real. We may as well absorb the $\alpha$ into $A$ and take $\beta = 1$. So we want a positive definite $S$ with $S A^T + A S = P$, where $P = B^T B$ is positive definite. Multiplying on left and right by $P^{-1/2}$ (the positive definite square root of $P^{-1}$), write this as $T C^T + C T = I$, where $T = P^{-1/2} S P^{-1/2}$ (which is positive definite iff $S$ is) and $C = P^{-1/2} A P^{1/2}$. Now this implies for every real vector $v$, $v^T C T v = v^T T C^T v = v^T v/2$. Of course if $C^T v = 0$ the left side is $0$, so this can't happen if $C$ is singular. If $C$ has a real eigenvalue $\lambda$, there is a nonzero vector $v$ such that $C^T v = \lambda v$, and then $\lambda v^T T v = v^T v/2$, so we get a contradiction if $\lambda < 0$. I suspect these are the only restrictions, i.e. if $C$ (and thus $A$) has no nonpositive real eigenvalues, there is a solution.
EDIT: My suspicion was false. Consider the case $ C = \pmatrix{1 & -3\cr 2 & -2\cr}$ which has two complex eigenvalues. The unique symmetric solution to $T C^T + C T= I$ is $ T = \left( \begin {array}{cc} -{\frac {17}{8}}&-{\frac {7}{8}}\\ -{\frac {7}{8}}&-{\frac {9}{8}}\end {array} \right) $ which of course is not positive definite.