Here's a statement of the Jacobson Density Theorem that I'm familiar with:
Let $R$ be a ring, $V$ a simple left $R$-module. By Schur, $\Delta := \operatorname{End}_R(V)$ is a division ring, and we can choose to have $\Delta$ act on $V$ on the right (Reverse Polish Notation style). Then by definition, $R$ acts on $V$ by $\Delta$-endomorphisms, inducing a map $$\phi: R \to \operatorname{End}_\Delta(V).$$ The density theorem says that for any $f \in \operatorname{End}_\Delta(V)$, and $v_1, \ldots, v_n \in V$, there exists $r \in R$ such that $r \cdot v_i = f(v_i)$ for each $i = 1, \ldots, n$.
Does this theorem require the assumption that $R$ is unital?
I think it is probably true without identity, and the resource to confirm this would be Jacobson's Structure of rings.
Unfortunately, I don't have a copy, but Jacobson did write this book meticulously based on rings not necessarily having identity, and chapter II goes into fiendish detail on versions of density theorems.
If a careful check of Jacobson's assumptions on the ring $\mathfrak{A}$ confirms he does not assume identity, then you can rest assured.