Let the excircle with center $J_A$ touch $BC$ at $F$, $P$ be the midpoint of the smaller arc $BC$ of the circumcircle of $ABC$, and $M$ be the midpoint of the side $BC$. Then the point $P$ is the midpoint of the line segment $IJ_A$ and hence the area of the triangle $IEP$ is half the area of the triangle $IEB$. Thus, the areas of the triangles $IEP$ and $IEO$ are equal. Hence the line $EI$ bisects the line segment $OP$. Let $Q$ be the midpoint of the line segment $EJ_A$ and $N$ be the midpoint of the line segment $OP$.

If $ED = AC - AB$, then $ED = DF = 2 \cdot DM$ and $EI = 2 \cdot IN$. Hence $I$ is the centroid of the triangle $OEP$, and since $D$ is the midpoint of the line segment $EF$, we conclude that the points $I$, $D$, $Q$ are collinear and $2 \cdot DQ = J_AF = r_a$. Since $PQ$ is parallel to $IN$ we get that $INPQ$ is a parallelogram. Hence
$$r + \frac{r_a}{2} = ID + DQ = IQ = NP = \frac{OP}{2} = \frac{R}{2}$$
and $R = 2r + r_a$.

If $R = 2r + r_a$, then let the line passing through $Q$ and perpendicular to $BC$ intersect the lines $BC$ and $EN$ at the points $D'$ and $I'$, respectively. Since $PQ$ is parallel to $I'N$, we obtain that $I'NPQ$ is a parallelogram and hence
$$I'Q = NP = \frac{R}{2} = r + \frac{r_a}{2}.$$
Since $D'Q$ is parallel to $J_AF$, we obtain that
$$D'Q = \frac{r_a}{2} = I'Q - r$$
and hence $D'I' = r = DI$. Therefore $D \equiv D'$ and $I \equiv I'$ which shows that $ED = DF = AC - AB$.