2 pointed out an error.

Yes, the map is injective.

In general, suppose I have a Lie algebra $\mathfrak g$ acting on a space $X$; then I get a Lie algebra structure on the vector bundle $\mathfrak g \times X$ with anchor map $\rho: (\mathfrak g \times X) \to {\rm T}X$. I can form the universal enveloping algebroid for each: $\mathfrak U \rho: \mathfrak U(\mathfrak g \times X) \to \mathfrak U({\rm T}X)$. The universal enveloping algebroid of a Lie algebroid $A \to X$ is the $\mathcal O(X)$-algebra generated by $\Gamma(A)$ with commutation relations coming from the bracket $[,]: \Gamma(A) \otimes_k \Gamma(A)$ — note that $\Gamma(A)$ is a $k$-Lie algebra but not an $\mathcal O(X)$-Lie algebra.

Anyway, the trivialization gives a map $\mathfrak U \mathfrak g \hookrightarrow \mathfrak U(\mathfrak g \times X)$ by sending each element of $\mathfrak g$ to the corresponding constant section (in fact, it determines an isomorphism of $\mathcal O(X)$-modules but not of algebras $\mathfrak U(\mathfrak g \times X) \cong \mathfrak U \mathfrak g \otimes \mathcal O(X)$, with each tensorand embedding as a subalgebra).

On the other hand, $\mathfrak U({\rm T}X)$ is the algebra of differential operators on $X$, and your map $\mathfrak U \mathfrak g \to \mathfrak U({\rm T}X)$ factors through $\mathfrak U \rho: \mathfrak U(\mathfrak g \times X) \to \mathfrak U({\rm T}X)$.

Now suppose that for densely many $x \in X$, the action map $\mathfrak g \to {\rm T}_x X$ is injective. Then $\Gamma\rho: \Gamma(\mathfrak g \times X) \to \Gamma({\rm T}X)$ is injective. In particular, any relation imposed in the construction of $\mathfrak U({\rm T}X)$ pulls back to a relation in $\mathfrak U(\mathfrak g \times X)$. Letting $X = k^n$ and $\mathfrak g = \mathfrak{gl}_n$ finishes your question. Edit: No it doesn't: fiberwise, the map $\mathfrak{gl} \to {\rm T}k^n$ cannot be injective, just by dimensions! So my argument is broken. It might be salvageable, so I'll leave it up in case someone has a fix (if so, let me know and I'll switch the answer to CW), but I won't be fixing it today.

(I was going to claim that as soon as $\mathfrak g \to \Gamma({\rm T}X)$ was injective, then the answer would be yes, but in fact the "densely many $x$" requirement is important. For example, let $\mathfrak g = k^2 = \operatorname{span}(e_1,e_2)$ be the two-dimensional abelian Lie algebra, and pick $X$ with opens $U_1,U_2$ so that there are nonzero functions supported on one and not the other. Let the basis vector $e_i$ of $\mathfrak g$ act by some non-zero vector field supported entirely in $U_i$. This is certainly a representation. But at the level of universal enveloping algebra/oids, the product $e_1e_2$ is non-zero in $\mathfrak U\mathfrak g$ and $\mathfrak U(\mathfrak g \times X)$ but acts as the zero differential operator.)

Oh, a final comment: Of course, you should probably sprinkle the words "sheaf of" throughout the above exposition.

1

Yes, the map is injective.

In general, suppose I have a Lie algebra $\mathfrak g$ acting on a space $X$; then I get a Lie algebra structure on the vector bundle $\mathfrak g \times X$ with anchor map $\rho: (\mathfrak g \times X) \to {\rm T}X$. I can form the universal enveloping algebroid for each: $\mathfrak U \rho: \mathfrak U(\mathfrak g \times X) \to \mathfrak U({\rm T}X)$. The universal enveloping algebroid of a Lie algebroid $A \to X$ is the $\mathcal O(X)$-algebra generated by $\Gamma(A)$ with commutation relations coming from the bracket $[,]: \Gamma(A) \otimes_k \Gamma(A)$ — note that $\Gamma(A)$ is a $k$-Lie algebra but not an $\mathcal O(X)$-Lie algebra.

Anyway, the trivialization gives a map $\mathfrak U \mathfrak g \hookrightarrow \mathfrak U(\mathfrak g \times X)$ by sending each element of $\mathfrak g$ to the corresponding constant section (in fact, it determines an isomorphism of $\mathcal O(X)$-modules but not of algebras $\mathfrak U(\mathfrak g \times X) \cong \mathfrak U \mathfrak g \otimes \mathcal O(X)$, with each tensorand embedding as a subalgebra).

On the other hand, $\mathfrak U({\rm T}X)$ is the algebra of differential operators on $X$, and your map $\mathfrak U \mathfrak g \to \mathfrak U({\rm T}X)$ factors through $\mathfrak U \rho: \mathfrak U(\mathfrak g \times X) \to \mathfrak U({\rm T}X)$.

Now suppose that for densely many $x \in X$, the action map $\mathfrak g \to {\rm T}_x X$ is injective. Then $\Gamma\rho: \Gamma(\mathfrak g \times X) \to \Gamma({\rm T}X)$ is injective. In particular, any relation imposed in the construction of $\mathfrak U({\rm T}X)$ pulls back to a relation in $\mathfrak U(\mathfrak g \times X)$. Letting $X = k^n$ and $\mathfrak g = \mathfrak{gl}_n$ finishes your question.

(I was going to claim that as soon as $\mathfrak g \to \Gamma({\rm T}X)$ was injective, then the answer would be yes, but in fact the "densely many $x$" requirement is important. For example, let $\mathfrak g = k^2 = \operatorname{span}(e_1,e_2)$ be the two-dimensional abelian Lie algebra, and pick $X$ with opens $U_1,U_2$ so that there are nonzero functions supported on one and not the other. Let the basis vector $e_i$ of $\mathfrak g$ act by some non-zero vector field supported entirely in $U_i$. This is certainly a representation. But at the level of universal enveloping algebra/oids, the product $e_1e_2$ is non-zero in $\mathfrak U\mathfrak g$ and $\mathfrak U(\mathfrak g \times X)$ but acts as the zero differential operator.)

Oh, a final comment: Of course, you should probably sprinkle the words "sheaf of" throughout the above exposition.