This is the theorem, which is taken from A Course in Computational Algebaic Number Theory, by Henri Cohen:
Ultimately, what I need is to understand what the "simple algorithm" is.
I think I understand clearly everything up to "the general solution is". From there, my understanding of the argument is as follows, with all polynomials expressed in $\mathbb{F}_p[X]$: $U$ and $V$ are Bézout coefficients for $A_e$ and $B_e$. Given a pair, $(U, V)$, of Bézout coefficients, all other pairs are of the form $$(U-KB_e/GCD(A_e,B_e),V+KA_e/GCD(A_e,B_e))$$ where $K$ is any polynomial in $\mathbb{F}_p[X]$. $UA_e+VB_e=1$ implies that $GCD(A_e,B_e)=1$, which simplifies the form of the Bézout coefficients to $$(U-KB_e,V+KA_e)$$ So for $L=U-KB_e$ and $M=V+KA_e$, $LA_e+MB_e=1$, and $DLA_e+DMB_e=D$. Taking $W=DK$, $DM$ and $DL$ give the same expressions for $S$ and $T$ as Cohen's proof does.
If my reasoning is right, it seems to imply that any choice of $W$ that is divisible by $D$, including $W=0$, (rather than only one particular choice as the wording of the proof suggested to me) yields an $S$ and a $T$ that obey the required congruences. Is that true? So the algorithm is that $S=VD$ and $T=UD$, so $A_{e+1}=A_e+p^eVD$ and $B_{e+1}=B_e+p^eUD$?