$\def\ZZ{\mathbb{Z}}\def\CC{\mathbb{C}}$ Let $E$ be an elliptic curve with affine chart $\{ (x,y) : y^2 = x+ax^2+bx^3 \}$ . We will write $p$ for the point $(0,0)$ and $\infty$ for the puncture. We want a holomorphic function on $E$ with a simple zero at $p$. We'll write $\omega$ for the nonvanishing holomorphic form
$$\omega = \frac{dx}{2y} = \frac{dy}{1+2ax+3 bx^2}$$
on $E$. If we think in terms of the universal cover $\phi: \CC \to E$, then $\phi^{\ast} x$, $\phi^{\ast} y$ and $\phi^{\ast} \omega$ are $\wp(z)$, $\wp'(z)$ and $dz$ respectively (up to scalar factors). I'll switch between $E$ and the universal cover $\CC$ as seems convenient.
Let $\gamma_1$ and $\gamma_2$ be a basis of cycles for $H_1(E)$, avoiding $p$ and $\infty$. Let $\delta$ be a path from $p$ to $\infty$, avoiding $\gamma_1$ and $\gamma_2$.
Claim There are constants $e$ and $f$ so that
$$\int_{\gamma_1} \left( \frac{y}{x} + e + f x \right) \omega = \int_{\gamma_2} \left( \frac{y}{x} + e + f x \right) \omega = 0 \quad (\ast).$$
Proof We need to know that the vectors $(\int_{\gamma_1} \omega, \int_{\gamma_2} \omega)$ and $(\int_{\gamma_1} x \omega, \int_{\gamma_2} x \omega)$ are linearly independent. In any particular example, this should be checkable by hand. The conceptual explanation is as follows:
On any smooth affine variety $X$, the closed algebraic $k$ forms represent all classes in $H^k_{DR}(X, \CC)$. In particular, there are algebraic $1$-forms on $E \setminus \{ \infty \}$ representing linearly independent forms $H_1(E \setminus \{ \infty \}) \to \CC$.
Now, $(1, \wp, \wp', \wp'', \wp''', \ldots)$ forms a basis for the coordinate ring of $E \setminus \{ \infty \}$. So $(dz, \wp dz, d \wp, d \wp', d \wp'', \ldots)$ forms a basis for algebraic $1$-forms on $E \setminus \{ \infty \}$. All of these but the first $2$ are exact, so the integrals of $dz$ and $\wp dz$ must span $\mathrm{Hom}(H_1(E \setminus \{ \infty \}), \CC)$. Translate back to the algebraic notation gives the claim. $\square$
Choose a base point $u$ in $E \setminus \delta$ and define
$$g(u) = \int_v^u \left( \frac{y}{x} + e + f x \right) \omega$$
where the integral is taken over any path in $E \setminus \delta$. Since we selected $e$ and $f$ to obey $(\ast)$, the integral is well defined, and has a branch discontinuity along $\delta$.
If $\rho$ is any closed curve around $p$, missing the $\gamma_i$ and transverse to $\delta$, then we have
$$\oint_{\rho} \left( \frac{y}{x} + a + b x \right) \omega = \oint \left( \frac{y}{y^2+\cdots} + \cdots \right)\left( \frac{dy}{1+\cdots} \right) = 2 \pi i$$
where the ellipses are holomorphic at $p$. So $g$ has a jump discontinuity of $2 \pi i$ over $\delta$.
Define $G(u) = \exp(g(u))$. Then $G(u)$ is holomorphic on all of $E \setminus \{ \infty \}$. It has a simple zero at $p$, an essential singularity at $\infty$ (from integrating $x \omega$ to get a pole of order $1$ and then exponentiating it) and is nonzero everywhere else.
In summary Find a $1$-form $\alpha$ with a residue of $2 \pi i$ at $p$, any pole you want at $\infty$ and integrals in $(2 \pi i) \ZZ$ over $H_1(E \setminus \{ p, \infty \})$. Then $\int \alpha$ is well defined on $E$ except for a branch cut from $p$ to $\infty$, and has discontinuity $2 \pi i$ over the branch cut. And $\exp(\int \alpha)$ is well defined on $E \setminus \{ \infty \}$, with a simple zero at $p$, essential singularity at $\infty$ and no other zeroes or poles. Found by reasoning backwards about what $dG/G$ should look like.
