Consider the simpler problem of two parametrized curves $(x_i(t),y_i(t))$ that start at $(0,0)$ at time $t=0$, and equality of trajectories up to reparametrization, which is stronger than equality of point sets and a more natural condition as it is local (up to matching of starting points). Heuristically, and to some extent rigorously, there is a usable criterion.
We want that when $x_1 = x_2$, then $y_1 = y_2$ so that the bivariate function $x_1(t)-x_2(s)$ divides $y_1(t)-y_2(s)$ (and vice versa) in a suitable ring of functions. Their ratio is an invertible function with positive values, at least for nonzero $s,t$ near $0$. In fact we need it to be positive only for nonzero $(s,t)$ at which $x_1(s)=x_2(t)$ or the same for $y$.
Example: parabola and half-parabola.
Curve A is $(t,t^2)$.
Curve B is $(s^2,s^4)$.
$x_1(t)-x_2(s) = t - s^2$
$y_1(t)-y_2(s) = t^2 - s^4$
Ratio is $(t + s^2)$
This is positive near the locus where $x_1(t)=x_2(s)$ (namely $t=s^2$).
On the locus where $y_1(t)=y_2(s)$ (namely $t^2=s^4$), this is positive for $t>0$ and negative for $t < 0$. The positivity condition knows which half of the parabola is curve B! That is a good sign that this is either the complete answer to the simplified problem, or on the right track.
Finding an intersection point of two parametric curves OR detecting a difference between the curves is simpler than the general problem of curve intersection. Take a point on one curve, solve for the parameter values that would place its $x$ coordinate on the other curve, and test whether the $y$ coordinates are the same. For algebraic parameterizations this calculation can be done exactly.
For the point-set equality problem, locate the zeros of the $st$ ratio. These parameter values segment the two curves into arcs. Then there is a combinatorial problem of orienting and matching (by the procedure given above) identical arcs of the two curves, and testing whether both curves are covered by the matched arcs.
The general equation for the tangent line to $f$ at the point $c$ is
$$y = f'(c) (x - c) + f(c)$$
Since the derivative of $e^{ax}$ is $a e^{ax}$, this takes the form
$$y = ae^{ac} (x - c) + e^{ac}$$
This needs to pass through the origin, so we need
$$0 = -ca e^{ac} + e^{ac} \implies e^{ac} (1 - ac) = 0$$
Since $e^{ac} \ne 0$ for any choice of $c$, we see that $c = 1/a$.
Best Answer
It's not so complicated!
You can assume the inflection point of the cubic at the origin. Up to similarity this means that we look at the cubic $$x\mapsto f(x):=x^3 - ax,\qquad a>0\ .$$ Let $\bigl(u,f(u)\bigr)$ be an arbitrary point on the cubic. The tangent there is then given by $$x\mapsto t(x)=f(u)+f'(u)(x-u)=(3u^2-a)x-2u^3\ .$$ One easily computes $$f(x)-t(x)=(u-x)^2(2u+x)\ .$$ This shows that the tangent intersects the cubic at the point $\bigl(-2u,f(-2u)\bigr)$. We therefore have to solve the equation $$f'(u)f'(-2u)+1=(3u^2-a)(12u^2-a)+1=0\ .$$ When $a>{4\over3}$ we have the four real solutions $$u=\pm\sqrt{{5a\pm\sqrt{9a^2-16}\over24}}\ .$$
See the Fig. below for $a=3/2$