It is enough to show xy generates x, and hence xR, whenever xy is nonzero. Thus there is no nonzero proper right submodule of xR.
Since 0 is a semiprime ideal, and xy is nonzero, there is some z such that xyzxy is nonzero. Let f be a left R module homomorphism on R, given by right multiplication by yzx. Remember that Rx is simple, so an endomorphism on Rx is trivial, or it is an automorphism. Apply f to Rx and get Rxyzx. Since 1xyzxy is nonzero, 1xyzx is nonzero, and f maps Rx onto Rx.
Let g be the inverse of f. (You can verify g is a left module homomorphism.) Thus x = g(f(x)) = g(xyzx) = xyg(zx), and xy generates x on the right. That completes the proof.
If R is not semiprime, a minimal left ideal need not generate a minimal right ideal. Let R be the upper triangular 2×2 matrices over a division ring K. Matrices with the right column zero form a left ideal, generated by [1,0|0,0]. This is a one dimensional K vector space, and a minimal left ideal. The right ideal generated by [1,0|0,0] consists of matrices with bottom row 0. This is a two dimensional K vector space, and it contains a smaller ideal, generated by [0,1|0,0]. This smaller right ideal has square equal to 0, showing the ring is not semiprime.