The plane curve $f(x,y)=x^2(x+1)-y^2=0\subset \Bbb{C}^2$ doesn't have a cusp at $(0,0)$ because we have a parametrization $(x,x\sqrt{1+x}),(x,-x\sqrt{1+x})$ of two distinct "analytic curves" around $(0,0)$.
In contrary $x^2-y^3=0$ has a cusp because $(y\sqrt{y},y)$ is not an analytic curve around $(0,0)$.
Over $\Bbb{C}$ we have the ring of all functions that are analytic in some neighborhood of $x_0=0$, but this doesn't generalize well to arbitrary fields.
We may try to replace it by the ring of formal power series in $x$, which is $k[[x]]=\varprojlim k[x]/(x^n)$,
or in $y$, which is $k[[y]]=\varprojlim k[y]/(y^n)$, but
it appears that the most insightful ring is $\varprojlim k[x,y]/(f(x,y))/(x,y)^n=k[[x,y]]/(f(x,y))$.
If $char(k)\ne 2$ then $\sqrt{1+x}=\sum_{k\ge 0} {1/2\choose k}x^k$ is in this ring and $(y-x\sqrt{1+x}),(y+x\sqrt{1+x})$ are two distinct prime ideals. $k[[x,y]]/(f(x,y))$ is not an integral domain as $(y-x\sqrt{1+x})(y+x\sqrt{1+x})=0$. In contrary to $k[[x,y]]/(x^2-y^3)$ which is an integral domain.
For an arbitrary variety this generalizes to $\varprojlim O_P / \mathfrak{p}^n$ where $O_P$ is the ring of rational functions regular at $P$ (or the localization thereof for a non-closed point) and $\mathfrak{p}$ is the prime ideal of those vanishing at $P$ (its unique maximal ideal).
For a curve, when a parametrization $(x,h(x))$ exists with $h $ a formal power series then in fact $h$ is algebraic over $k[x]$ so that it can be seen as a rational function on a ramified covering of our curve.
Assume for convinience that P is the origin in some chart and write the equation of $F_*$ as:
$$
F_*=\sum_{i,j}a_{ij}x^iy^j,
$$
The tangent lines will be the zeros of the polynomial:
$$
T=\sum_{i+j=m} a_{ij}x^iy^j,
$$
Where $m$ is the multiplicity of $F$ at $P$.
Notice that if $L=bx+cy$ divides $T$, then $I(P,F\cap L)>m$.
Best Answer
The key here is the points at infinity: your desired equation has an inflection point at infinity, while your current equation may not have this (at least if $bx^3+cx^2y+dxy^2+ey^3$ is not a cube). So you need to set things up so the portion of your curve at infinity is an inflection point.
In general, the way to start with this is to first find the inflection point via the Hessian: given an irreducible projective plane curve cut out by an equation $f(x,y,z)=0$ of degree $d$, if $d-1$ is coprime to the field characteristic, the common vanishing locus of $f$ and the Hessian determinant of $f$ contains all the inflection points on $V(f)$ (see Fulton's Algebraic Curves for full proof, for instance). It turns out that for an irreducible plane cubic over a field of characteristic not $2$, one will always get at least one nonsingular inflection point out of this, and one can then choose coordinates so that point is $[0:1:0]$.
From there, apply the techniques you've mentioned in the post: you'll find that when you substitute in $Z=0$, you must have your equation become $X^3$, so $c=d=e=0$. Now all you have to do is turn $aY^2Z-bX^3$ in to $Y^2Z-X^3$, which you can do via a clever linear scaling of $X$ and $Y$, which I give an explicit form of below in the spoiler.
For a proof of the theorem that there are exactly two singular projective plane cubics up to projective equivalence which works in all characteristics, one may prove that any singular projective plane cubic is a projection of the twisted cubic in $\Bbb P^3$ from a point not on the curve. Projection from points on the tangent variety give cuspidal cubics, while projection from points on the the secant variety gives nodal cubics. I don't have a precise reference for this at present, but I think it should be in one of Harris' books. (One may also mess around with the pseudo-Hessian, a gadget designed to make the Hessian work a bit better in characteristic two, but this is a bit more obscure.)