[proofplan] We use the full semistable modularity theorem of Wiles and Taylor-Wiles as the external modularity-lifting input: for every prime $\ell$, the $\ell$-adic Galois representation attached to a semistable elliptic curve over $\mathbb Q$ is realized by a normalized weight $2$ newform of exact level equal to the conductor, with matching local factors at all primes. The good-prime equality identifies Hecke eigenvalues with Frobenius traces, while the bad-prime compatibility identifies the $U_p$ eigenvalue with the split or nonsplit multiplicative reduction sign. Equality of the local Euler factors gives equality of the Euler products, hence $L(E,s)=L(f,s)$. [/proofplan]

[step:Invoke the semistable modularity theorem with the conductor of $E$] Let $E/\mathbb Q$ be the semistable elliptic curve in the statement, and let $N$ be its conductor. Let $\ell$ denote an arbitrary prime number, and let \begin{align*} \rho_{E,\ell}: \operatorname{Gal}(\overline{\mathbb Q}/\mathbb Q) &\to GL_2(\mathbb Q_\ell) \end{align*} denote the $\ell$-adic Galois representation on the rational Tate module of $E$. Let $\mathfrak H := \{z \in \mathbb C : \operatorname{Im}(z) > 0\}$ denote the complex upper half-plane. We use the following full external theorem of Wiles and Taylor-Wiles: if $E/\mathbb Q$ is semistable of conductor $N$, then the compatible system $(\rho_{E,\ell})_\ell$ is attached to a normalized cuspidal Hecke eigenform \begin{align*} f: \mathfrak H &\to \mathbb C \end{align*} of weight $2$ and exact level $\Gamma_0(N)$, and the local Weil-Deligne representations, hence the local Euler factors, agree at every prime. In particular, the Hecke eigenvalues agree with the Frobenius traces of $E$ at every prime of good reduction, and the $U_p$ eigenvalue agrees with the multiplicative reduction sign at every prime $p \mid N$. Semistability is exactly the hypothesis required by this cited theorem, and the conductor appearing in the theorem is the same conductor $N$ appearing in the statement. [/step]

[step:Choose the normalized newform and fix its Fourier notation] The full modularity theorem already produces a normalized newform of exact level $N$, so no further level-lowering argument is needed. For $z \in \mathfrak H$, define $q := e^{2\pi i z}$. We write the Fourier expansion of the normalized newform as \begin{align*} f(z) = \sum_{m=1}^{\infty} a_m(f) q^m, \end{align*} where $a_m(f) \in \mathbb C$ denotes the $m$th Fourier coefficient of $f$. The normalization condition is $a_1(f)=1$, and $f \in S_2(\Gamma_0(N))$ by the theorem just invoked. [/step]

[step:Identify the Euler factors prime by prime] For each prime $p$, let $L_p(E,s)$ denote the local Euler factor of $E$ at $p$, and let $L_p(f,s)$ denote the local Euler factor of the newform $f$ at $p$. If $p \nmid N$, then $E$ has good reduction at $p$ and the modularity theorem gives \begin{align*} a_p(f) = p + 1 - \#E(\mathbb F_p). \end{align*} Therefore \begin{align*} L_p(E,s) = \left(1 - a_p(f)p^{-s} + p^{1-2s}\right)^{-1} = L_p(f,s). \end{align*} If $p \mid N$, semistability means that the reduction of $E$ at $p$ is multiplicative. Define \begin{align*} \varepsilon_p(E) := \begin{cases} 1, & \text{if the multiplicative reduction at } p \text{ is split},\\ -1, & \text{if the multiplicative reduction at } p \text{ is nonsplit}. \end{cases} \end{align*} The bad local Euler factor of $E$ is then \begin{align*} L_p(E,s) = \left(1 - \varepsilon_p(E)p^{-s}\right)^{-1}. \end{align*} For the newform $f$, let $a_p(f)$ denote its $U_p$ eigenvalue at the prime $p \mid N$; the bad local Euler factor of $f$ is \begin{align*} L_p(f,s) = \left(1 - a_p(f)p^{-s}\right)^{-1}. \end{align*} The local compatibility in the Wiles-Taylor-Wiles theorem gives $a_p(f)=\varepsilon_p(E)$, so again \begin{align*} L_p(E,s) = L_p(f,s). \end{align*} Thus the local Euler factors agree for every prime $p$. [/step]

[step:Conclude equality of the global $L$-functions] The Hasse-Weil $L$-function of $E$ and the Hecke $L$-function of $f$ are defined by their Euler products \begin{align*} L(E,s) &= \prod_p L_p(E,s), & L(f,s) &= \prod_p L_p(f,s), \end{align*} which converge absolutely for $\operatorname{Re}(s)$ sufficiently large. Since $L_p(E,s)=L_p(f,s)$ for every prime $p$, the two Euler products are equal on that half-plane. By [analytic continuation](/page/Analytic%20Continuation) of the modular $L$-function and the equality already obtained in its domain of absolute convergence, this gives \begin{align*} L(E,s)=L(f,s). \end{align*} Hence there exists a weight $2$ newform $f \in S_2(\Gamma_0(N))$ whose $L$-function agrees with $L(E,s)$, as required. [/step]