= Solution
Extend the <Dirichlet character> $\chi$ periodically to all <integers>, putting $\chi(n)=0$ when $(n,N)>1$. Its <Dirichlet L-function>, initially on $\operatorname{Re}s>1$, is
$$
\boxed{L(\chi,s)=\sum_{n\geq1}\chi(n)n^{-s}
=\prod_{p\nmid N}(1-\chi(p)p^{-s})^{-1}.}
$$
The <Euler product> follows from unique <prime factorization> and <absolute convergence>; a <Dirichlet character> is completely multiplicative on this extension. Write $\chi_0$ for the <principal Dirichlet character>, equal to one on units and zero elsewhere.
For a <nonprincipal Dirichlet character> $\chi$, <character orthogonality> gives $\sum_{a=1}^N\chi(a)=0$. Explicitly, multiplication by a unit $u$ with $\chi(u)\ne1$ permutes the unit residues and multiplies this sum by $\chi(u)$, forcing it to vanish. For $t>0$ set
$$
H_\chi(t)=\sum_{n\geq1}\chi(n)e^{-nt}
=\frac{\sum_{a=1}^N\chi(a)e^{-at}}{1-e^{-Nt}}.
$$
The numerator is $O(t)$ at zero and the denominator is $Nt+O(t^2)$, so $H_\chi$ is bounded, indeed analytic, near zero; it decays exponentially at infinity. Initially for $\operatorname{Re}s>1$, <absolute convergence> justifies
$$
\Gamma(s)L(\chi,s)=\int_0^\infty H_\chi(t)t^{s-1}\,dt.
$$
This <Mellin transform> integral is holomorphic for $\operatorname{Re}s>0$, locally uniformly in $s$, and division by the <Gamma function> proves the requested <analytic continuation> to the left of the line one.
In fact, the same argument proves <Mellin continuation of a nonprincipal Dirichlet L-function> to the entire plane. If $H_\chi(t)=\sum_{r=0}^M c_rt^r+O(t^{M+1})$ at zero, subtract this <Taylor polynomial> on $(0,1)$ and add its explicit integrals:
$$
\Gamma(s)L(\chi,s)=\int_1^\infty H_\chi(t)t^{s-1}\,dt
+\sum_{r=0}^M\frac{c_r}{s+r}
+\int_0^1\left(H_\chi(t)-\sum_{r=0}^M c_rt^r\right)t^{s-1}\,dt.
$$
The last integral is holomorphic on $\operatorname{Re}s>-M-1$. Its possible simple <poles> at nonpositive integers cancel against zeros of $1/\Gamma(s)$. Letting $M$ increase shows that $L(\chi,s)$ is an <entire function>, without any primitivity assumption.
For real $s>1$, use the absolutely convergent <Euler product> logarithm
$$
B_\chi(s)=\sum_{p\nmid N}\sum_{m\geq1}\frac{\chi(p)^m}{mp^{ms}}
=F_\chi(s)+R_\chi(s),\qquad e^{B_\chi(s)}=L(\chi,s).
$$
The higher-power remainder has the uniform estimate
$$
|R_\chi(s)|\leq\sum_p\sum_{m\geq2}p^{-m}
=\sum_p\frac1{p(p-1)}\leq\sum_{n\geq2}\frac1{n(n-1)}=1.
$$
For a <nonprincipal Dirichlet character> $\chi$, invoke the allowed <Nonvanishing of a nonprincipal Dirichlet L-function at one>. Its holomorphy and nonvanishing give a <holomorphic logarithm> on a small disk about one. On the connected real interval $1<s<1+\eta$, $B_\chi(s)$ differs from this logarithm by a fixed element of $2\pi i\mathbb Z$: the difference is continuous with exponential one. Thus the <prime-character sum near one> is bounded. This branch argument is needed for complex-valued <Dirichlet characters>.
For $\chi_0$,
$$
L(\chi_0,s)=\zeta(s)\prod_{p\mid N}(1-p^{-s}).
$$
The finite product has a positive limit as $s\downarrow1$, and the residue-one <pole> of the <Riemann zeta function> gives
$$
\boxed{F_{\chi_0}(s)=\log\frac1{s-1}+O(1)\longrightarrow+\infty,
\qquad F_\chi(s)=O(1)\quad(\chi\ne\chi_0).}
$$
Here $O(1)$ for nonprincipal <Dirichlet characters> means bounded complex magnitude.
Finally, for a <residue class> $a$ coprime to $N$, <Orthogonality of Dirichlet characters> gives
$$
\sum_{p\equiv a\pmod N}p^{-s}
=\frac1{\varphi(N)}\sum_{\chi\bmod N}\overline{\chi(a)}F_\chi(s)
=\frac1{\varphi(N)}\log\frac1{s-1}+O(1).
$$
Only <primes> not dividing $N$ occur, so the character orthogonality applies to every term. This sum diverges as $s\downarrow1$. A finite collection of <primes> would give a bounded sum, a contradiction. \b[Every reduced residue class contains infinitely many <primes>.] This is the <Dirichlet theorem on primes in arithmetic progressions>; the coprimality hypothesis is essential.
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