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Here is an elementary solution in a very special case. It assumes that the curve is already embedded in the plane using a Weierstrass-like equation, so you can't use it to later prove that every elliptic curve can be given by a Weierstrass equation, but perhaps it will still be instructive.

We assume the base field is of characteristic $0$ and algebraically closed. Suppose $E$ is the projective curve in $\mathbb{P}^2$ defined by the following equation in $\mathbb{A}^2$, $$y^2 = (x - \lambda_1)(x - \lambda_2)(x - \lambda_3)$$ where $\lambda_1, \lambda_2, \lambda_3$ are non-zero and distinct. It can be shown that $E$ is a smooth curve with a unique point at infinity, namely $P = (0 : 0 : 1)$. As Nils Matthes has indicated, to show that $\ell (n P) = n$ for all $n \ge 1$, it suffices to prove the case $n = 1$.

Suppose, for a contradiction, that $\ell (P) > 1$. We know that for all divisors $D$, $\ell (D) \le \deg D + 1$, so that means $\ell (P) = 2$, and so there is some non-constant rational function $t$ such that $L (P)$ is spanned by $1$ and $t$. Moreover, $t$ has a simple pole at $P$ (and nowhere else), so $t^n$ has a pole of order $n$ at $P$ (and nowhere else). As such, $L (n P)$ is spanned by $1, t, \ldots, t^n$. However, we know that $y$ is in $L (2 P)$ and $x$ is in $L (3 P)$, so that means, for some constants, \begin{align} x & = a_0 + a_1 t + a_2 t^2 \\ y & = b_0 + b_1 t + b_2 t^2 + b_3 t^3 \end{align} and by rescaling $t$ if necessary, we may assume that $a_2 = b_3 = 1$. (Note that $a_2^3 = b_3^2$.) We may also complete the square to assume that $a_1 = 0$. But then substituting into the original equation, we get $$(b_0 + b_1 t + b_2 t^2 + t^3)^2 = (t^2 + a_0 - \lambda_1) (t^2 + a_0 - \lambda_2) (t^2 + a_0 - \lambda_3)$$ which is impossible, since the RHS has six distinct zeros.