[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]
