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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?

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    You mean does R have to have a unit ? Then no it is not required. – 2017-01-20
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    @ReneSchipperus Yes, thanks! If you don't mind me asking, is there a different meaning of "unital"? That seems like something I should know if there is. – 2017-01-20
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    @DustanLevenstein It could have been interpreted as "shares the same unit as its superring" ($End_\Delta(V)$ in this case.) – 2017-01-20
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    @rschwieb I suppose so. – 2017-01-21

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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.

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    Thanks! I don't have a copy either, but I will check if it's at my school's library to be sure. – 2017-01-21