Try this.

$U$ is faithful and reflects invertibility, as in Beck's theorem.

For every "algebra" $A$, ie object of $\textbf C$,

there are a "set" $X$ and a "function" $\eta:X\to U A$ in $\textbf D$ that is universal in the sense that

for every other "algebra" $B$ and "function" $f:X\to U B$ in $\textbf D$ there is a unique "homomorphism" $h:A\to B$ in $\textbf C$ that extends $f$ in the sense that $f=\eta;U h$.

Try this. I confess I haven't written out the proofs yet, but will do so if you (and the community) think this is an appropriate answer. Please excuse my renaming your categories according to my personal convention.

First, $U:{\mathcal A}\to{\mathcal S}$ must be faithful and reflect invertibility, cf Beck's theorem.

We rewrite essential surjectivity of $F:{\mathcal S}\to{\mathcal A}$ in terms of univeral properties:

For every "algebra" $A$, ie object of $\mathcal A$,

there are a "set" $X$ and a "function" $e:X\to U A$ in $\mathcal S$ that is universal in the sense that

for every other "algebra" $B$ and "function" $f:X\to U B$ in $\mathcal S$ there is a unique "homomorphism" $h:A\to B$ in $\mathcal A$ that extends $f$ in the sense that $f=e;U h$.

How do we find the object $X$? This is going to invoke $F$ (if that's allowed by the question!)

It need not be unique (up to iso), but the construction of the replacement for $\mathcal S$ is essentially the one in this paper of mine and the ones by Hayo Thielecke and Peter Selinger that it cites.

If appropriate equalisers exist in $\mathcal S$ then the canonical $e:X\to U A$ (in fact the terminal one) is given by the equaliser of $U A\rightrightarrows U F U A$.

In terms of $X$, the two maps are $\eta U F X=\eta U A$ and $U F\eta X$. We can derive the second of these from the same problem starting with $F U A$ in place of $A$, in which case we get a split equaliser.