$A_n$ can be embedded in $SL_n(R)$, for any commutative ring R.
My thinking: For the case of $S_n$, the mapping $f:S_n \rightarrow GL_n(R)$ defined as $\sigma \mapsto [\sigma (e_1); \sigma (e_2);...; \sigma (e_n)]$
(where $I:=[e_1;e_2;...;e_n]$ is the identity matrix) serves as an isomorphism.
But I cannot find a way to put $A_n$ inside $SL_n(R)$.
You've already written the formula in your question: the embedding $f : S_n \to GL_n(R)$ defined by $\sigma \mapsto [\sigma(e_1);\sigma(e_2);...;\sigma(e_n)]$ restricts to an embedding $f : A_n \to SL_n(R)$.
The one thing that needs to be proved is that the image of each $\sigma \in A_n$ has determinant $+1$. When you factor $\sigma \in A_n$ as a product of an even number of transpositions $\sigma = \tau_1 \circ \cdots \circ \tau_{2k}$, each of the matrices $[\tau_i(e_1);\tau_i(e_2);...;\tau_i(e_n)]$ is obtained from the identity matrix by swapping two rows and hence has determinant $-1$. So the determinant of the product becomes $(-1)^{2k}=+1$.