First, cutting by the plane $z=d$ does not change things much. The resulting cone is similar to the bigger cone with the base in $z=0$. The similarity factor is $(c-d)/c$, so the area is $(c-d)^2/c^2$ times the area of the bigger cone.
So I'm going to consider the cone with the base the circle of radius $r$ centered at O, and the vertex at $(a,b,c)$. The base has area $\pi r^2$, of course. The question is what to do with the lateral surface. Here is the calculus approach (I can't think of another one, actually):
The base circle is parametrized by $(r\cos t, r\sin t,0)$. Using the equation of the line segment from this point to $(a,b,c)$, we arrive at the parametrization of the entire surface: $ \vec r(t,s)=\langle sa+(1-s)r\cos t, sb+(1-s)r\sin t, sc \rangle,\qquad 0\le s\le 1, \quad 0\le t\le 2\pi $ The area is the integral of $|\vec r_t\times \vec r_s|$. Let's do this: $\vec r_t= (1-s) \langle -r\sin t, r\cos t, 0 \rangle$ $\vec r_s=\langle a-r\cos t, b-r\sin t, c \rangle$ and the cross product simplifies to $\vec r_t\times \vec r_s = r(1-s)\langle c\cos t, c\sin t, r-a\cos t-b\sin t \rangle$ Integration with respect to $s$ yields a factor of $1/2$, hence the lateral area is $A=\frac{r}{2}\int_0^{2\pi}\sqrt{c^2+(r-a\cos t-b\sin t)^2}\,dt$ When $a=b=0$, the answer is $A=\pi r\sqrt{c^2+r^2}$. But when either of them is nonzero, we get a complete elliptic integral, which cannot be expressed in terms of elementary functions.
Finally, some informal reasoning to show that this outcome is not surprising. You can think of the skewed circular cone as a right cone with elliptical base. Recall that the perimeter of the base enters the computation of lateral area of a right circular cone. But the length of an ellipse is expressible only in terms of elliptic integrals -- which is why they are called elliptic integrals.