This is subtle. It is true that every elliptic curve can be written in this form, and that every smooth projective curve of this form has genus $1$.
But an elliptic curve is not a smooth projective curve of genus $1$. An elliptic curve (over a field $K$) is a smooth projective curve of genus $1$ together with a distinguished $K$-rational point, and this is key to the theory because it's the point that forms the identity for the group law. You need to say something about what this distinguished point is: with an equation of this form it's conventional to take it to be the point at infinity, with coordinates $(X : Y : Z) = (0 : 1 : 0)$. But this needs to be said explicitly in the definition; it is in fact possible to pick another point, which changes the group law. This is important for the following reasons among others:
There are smooth projective curves of genus $1$ over non-algebraically closed fields $K$ which have no $K$-rational points, and hence there is no choice of point which turns them into elliptic curves.
Elliptic curves have endomorphisms and automorphisms, and these are required to preserve the distinguished point (which turns out to imply that they preserve the group law). If you don't keep this in mind you will be confused when you read statements about endomorphisms and automorphisms of elliptic curves in the literature (which in fact happened recently on MO). For example, endomorphisms of an elliptic curve form a ring, but only if you require that endomorphisms preserve the distinguished point.
Just to get this off the unanswered list.
If $F$ is reducible (e.g. $F(x,y,z)=(x-y)(z-w)(y-z)$) then it cannot be smooth. The reason is the following nice observation:
Observation 1: Let $X$ be a Noetherian regular connected scheme. Then, $X$ is integral.
Proof: We need to show that $X$ is irreducible and reduced. Since $\mathcal{O}_{X,x}$ is regular for all $x$, and regular rings are integral (e.g see [1, Tag00NP]) we see that $X$ is clearly reduced. To see that it’s irreducible note that we can decompose $X$ into finitely many irreducible components $C_i$. Note then that if $X$ is not irreducible we must have that $C_i\cap C_j\ne \varnothing$ for some $i\ne j$ else then $\displaystyle X=\bigsqcup_i C_i$ which contradicts that $X$ is connected. But, let $x$ be a closed point of $C_i\cap C_j$ (for the $i,j$ where they intersect). Note then that $\mathcal{O}_{X,x}$ cannot be integral since $\mathrm{Spec}(\mathcal{O}_{X,x})$ contains $\eta_{C_i}$ which are distinct minimal primes. Thus, we arrive at a contradiction. Thus, $X$ is irreducible as desired. $\blacksquare$
This seems eminently useful if you know, a priori, that your hypersurface $E:=V(F)\subseteq \mathbb{P}^2_k$ is connected. But, again, this is actually automatic!
Observation 2: Let $F\in k[x,y,z]$ be a non-zero homogenous polynomial. Then, $C:=V(F)\subseteq \mathbb{P}^2_k$ is connected.
Proof: There are several proofs of this fact. But, one is to note that if $n=\deg(F)$ then we have a short exact sequence
$$0\to \mathcal{O}(-n)\to \mathcal{O}_{\mathbb{P}^2_k}\to i_\ast \mathcal{O}_C\to 0$$
where $i:C\hookrightarrow \mathbb{P}^2_k$ is the inclusion. From the LES in cohomology we see that we have an exact sequence
$$0\to \mathcal{O}_{\mathbb{P}^2_k}(\mathbb{P}^2_k)\to (i_\ast \mathcal{O}_{\mathbb{P}^2_k})(\mathbb{P}^2_k)\to H^1(\mathbb{P}^2_k,\mathcal{O}(-n))$$
But, this last group is zero (e.g. see [1, Tag01XT]) and so we see that
$$k\cong \mathcal{O}_{\mathbb{P}^2_k}(\mathbb{P}^2_k)\cong (i_\ast\mathcal{O}_C)(\mathbb{P}^2_k)=\mathcal{O}_C(C)$$
and so $\mathcal{O}_C(C)$ has no non-trivial idempotents, and so is connected. $\blacksquare$
EDIT: This is somewhat tangential, but alternatively to Observation 1 if you want to assume that your $E$ is a group, then it’s connected if and only if it’s irreducible (e.g. see Lemma 8 of this blog post of mine).
References:
[1] Stacks project authors. The Stacks Project. https://stacks.math.columbia.edu
Best Answer
This change of coordintes is linear, so it extends to a change of coordinates on $\Bbb P^2$ by $X\mapsto X$, $Y\mapsto (Y-a_1X-a_3Z)/2$, $Z\mapsto Z$. You can see directly that this preserves the point at infinity $[0:1:0]$, and that it gives you the projective equation $Y^2Z=4X^3+b_2X^2Z+2b_4XZ^2+b_6Z^3$.