Yes. Your transformation $\phi$ is an affine linear transformation, so it takes straight lines to straight lines, and therefore three points $P,Q,R$ are collinear if and only if $\phi(P),\phi(Q),\phi(R)$ are collinear. Since the group law is defined by the fact that $P+Q+R=0$ if and only if $P,Q,R$ are collinear, your result follows.
If you're following Silverman, a bit of a cheat sheet is as follows.
We have $E/K$ denoting the underlying elliptic curve- naively this is "$E$, which is cut out in some projective space by some equations, and a marked $K$-point $O$", or even more simply you could take it to be "the curve cut out in $\mathbb{P}^2$ by a Weierstrass equation with the understanding that $(0:1:0)$ is $O$".
Suppose that $L/K$ is an extension of fields. He uses $P \in E(L)$ to denote a $L$-rational point on $E$ (i.e., the defining equations for $E$ vanish at $P$, and the coordinates of $P$ are in $L$).
Where he says $P \in E$ he means that $P$ is a geometric point, i.e., $P \in E(\bar{K})$.
To address your confusion. You say "as I read more, I see an elliptic curve over $K$ comprises only of points over $K$". This is definitely not the case, that is the $K$-points on an elliptic curve over $K$. The curve itself "is" the equations and the point $O$ - which in turn describe an awful lot more information (e.g., the points over every extension $L/K$).
Of course, different authors will have different conventions with respect to these things.
Best Answer
An elliptic curve over a field $F$, in fact, is a projective, smooth curve of genus $1$, with at least one point defined over $F$. It turns out (as Lord Shark discusses) there is a change of variables that brings any elliptic curve to a model of the form you write (which is called a Weierstrass equation). In fact, if the characteristic of $F$ is not $2$ or $3$, then you can bring it to a model of the form $y^2=x^3+Ax+B$, which is called a short Weierstrass form.
For instance, the curve $C$ over $\mathbb{Q}$ given by $x^3+y^3=1$ (the curve $X^3+Y^3=Z^3$ in projective space) is also an elliptic curve (there is at least one point, namely $(1,-1)$, the point $[1,-1,0]$ in projective coordinates), even though it is not given a priori by a Weierstrass form. A change of variables brings $C$ to the equation $$y^2 - 9y = x^3 - 27$$ in Weierstrass form.
If you are interested in how one finds such changes of variables, this is briefly explained in Silverman and Tate's "Rational Points on Elliptic Curves".