The diagonal entries of $D$ are the reciprocals of the row sums of $A$. The row sums of $B$ are those of $A$. Thus $D$ is known. Then $A$ can be obtained as
$A=\frac1{\sqrt D}\sqrt{\sqrt DB\sqrt D}\frac1{\sqrt D}\;,$
or, if you prefer,
$A=D^{-1/2}\left(D^{1/2}BD^{1/2}\right)^{1/2}D^{-1/2}\;.$
According to this post, this is the unique symmetric positive-definite solution of $ADA=B$.
The square root of $D$ is straightforward; the remaining square root can be computed by diagonalization or by various other methods.
To see that the solution is consistent in that the $A$ so obtained does indeed have the same row sums as $B$, note that
$\left(D^{1/2}BD^{1/2}\right)\left(D^{-1/2}\mathbf 1\right)=D^{1/2}B\mathbf 1=D^{1/2}D^{-1}\mathbf 1=D^{-1/2}\mathbf 1\;,$
where $\mathbf 1$ is the vector consisting entirely of $1$s. Thus $D^{-1/2}\mathbf 1$ is an eigenvector with eigenvalue $1$ of $D^{1/2}BD^{1/2}$, and thus also of
$D^{1/2}AD^{1/2}=\left(D^{1/2}BD^{1/2}\right)^{1/2},$
and thus
$DA\mathbf1=D^{1/2}\left(D^{1/2}AD^{1/2}\right)\left(D^{-1/2}\mathbf1\right)=D^{1/2}D^{-1/2}\mathbf1=\mathbf1$
as desired.
Perhaps a more concise way of saying all this is that we should apply a transform x=D^{1/2}x' to get
x^\top Ax=x'^\top A'x'\quad\text{with}\quad A'=D^{1/2}AD^{1/2}\;,\\ x^\top Bx=x'^\top B'x'\quad\text{with}\quad B'=D^{1/2}BD^{1/2}\;,\\ \mathbf1'=D^{-1/2}\mathbf1\;,
and then the equation becomes A'^2=B' and the row sum conditions become A'\mathbf1'=B'\mathbf1'=\mathbf1'.