Is my $A_0$ correct?
No, it is not correct. Let $D,E,F$ be the tangent point of the inscribed circle with the side $BC,CA,AB$ respectively. Then, noting that
$$AE=AF=s-a,\quad BD=BF=s-b,\quad CE=CD=s-c$$
where $s=(a+b+c)/2$, we have
$$\begin{align}A_0&=[\triangle{ABC}]-([\triangle{AEF}]+[\triangle{BDF}]+[\triangle{CDE}])\\&=A-\frac 12(s-a)^2\sin A-\frac 12(s-b)^2\sin B-\frac 12(s-c)^2\sin C\end{align}\tag1$$
By the way, let $K,L,M$ be the tangent point of the escribed circle in $\angle A$ with the side $BC,CA,AB$ respectively. Then, noting that
$$CK=CL=s-b,\quad BK=BM=s-c,$$
we have
$$\begin{align}[\triangle{KLM}]&=[\triangle{AML}]-([\triangle{ABC}]+[\triangle{BKM}]+[\triangle{CKL}])\\&=\frac 12s^2\sin A-A-\frac 12(s-c)^2\sin(\pi-B)-\frac 12(s-b)^2\sin(\pi-C)\\&=\frac 12s^2\sin A-A-\frac 12(s-c)^2\sin B-\frac 12(s-b)^2\sin C\end{align}$$
Similarly, the areas of the other triangles formed with escribed circles are given by
$$\frac 12s^2\sin B-A-\frac 12(s-a)^2\sin C-\frac 12(s-c)^2\sin A$$
$$\frac 12s^2\sin C-A-\frac 12(s-a)^2\sin B-\frac 12(s-b)^2\sin A$$
Hence, from $(1)$, we have
$$2A+A_0-A_1-A_2-A_3$$$$=2A+A-\frac 12(s-a)^2\sin A-\frac 12(s-b)^2\sin B-\frac 12(s-c)^2\sin C-(\frac 12s^2\sin A-A-\frac 12(s-c)^2\sin B-\frac 12(s-b)^2\sin C)-(\frac 12s^2\sin B-A-\frac 12(s-a)^2\sin C-\frac 12(s-c)^2\sin A)-(\frac 12s^2\sin C-A-\frac 12(s-a)^2\sin B-\frac 12(s-b)^2\sin A)$$$$=6A+\frac 12\sin A(-(s-a)^2-s^2+(s-c)^2+(s-b)^2)$$$$+\frac 12\sin B(-(s-b)^2+(s-c)^2-s^2+(s-a)^2)$$$$+\frac 12\sin C(-(s-c)^2+(s-b)^2+(s-a)^2-s^2)$$$$=6A+\frac 12\sin A(-2bc)+\frac 12\sin B(-2ca)+\frac 12\sin C(-2ab)=6A-2A-2A-2A=0$$