Here are two equivalences.

**Theorem 1**. For each $m \geq 1$, the following are equivalent.

a) For all nontrivial Dirichlet characters $\chi \bmod m$, $L(1,\chi) \not= 0$.

b) For all $a \in (\mathbf Z/m\mathbf Z)^\times$, the set of primes $p \equiv a \bmod m$ has Dirichlet density $1/\varphi(m)$.

*Proof of Theorem 1*.
We will will compute the Dirichlet density of $\{p \equiv a \bmod m\}$ without assuming (a) and then see why (a) and (b) are equivalent. The trivial Dirichlet character modulo $m$ will be written as $\mathbf 1_m$.

For each Dirichlet character $\chi \bmod m$, $L(s,\chi)$ is analytic for ${\rm Re}(s) > 0$ except for $L(s,{\mathbf 1}_m)$ having a simple pole at $s = 1$. Set
$$
n(\chi) := {\rm ord}_{s=1}(L(s,\chi))
$$
so $n({\mathbf 1}_m) = -1$ and $n(\chi) \geq 0$ for all nontrivial $\chi$.

For ${\rm Re}(s) > 1$ and $(a,m) = 1$,
$$
\sum_{p \equiv a \bmod m} \frac{1}{p^s} =
\frac{1}{\varphi(m)} \sum_{p}\sum_{\chi} \frac{\chi(p)\overline{\chi}(a)}
{p^s} =
\frac{1}{\varphi(m)} \sum_{\chi}\overline{\chi}(a)\left(\sum_{p} \frac{\chi(p)}
{p^s}\right)
$$
where the sum on the right run over all primes $p$ and all Dirichlet characters $\chi \bmod m$.
For a Dirichlet character $\chi \bmod m$ and ${\rm Re}(s) > 1$,
$$
\log L(s,\chi) = \sum_p \frac{\chi(p)}{p^s} + \sum_{p,k\geq 2} \frac{\chi(p^k)}{kp^{ks}} = \sum_p \frac{\chi(p)}{p^s} + O(1),
$$
where the $O$-constant is $\sum_{p,k \geq 2} 1/(kp^k)$, so
$$
\sum_{p \equiv a \bmod m} \frac{1}{p^s} =
\frac{1}{\varphi(m)}\sum_{\chi \bmod m} \overline{\chi}(a)\log L(s,\chi) + O(1).
$$

Now let's bring in the order of vanishing $n(\chi)$ above. For $s$ near $1$, $L(s,\chi) = (s-1)^{n(\chi)}f_\chi(s)$ where $f_\chi(s)$ is an analytic function in a neighborhood of $s = 1$ and $f_\chi(1) \not= 0$. Therefore $f_\chi(s)$ has an analytic logarithm around $s = 1$ (well-defined up to adding an integer multiple of $2\pi i$), so for $s > 1$,
$\log L(s,\chi) = n(\chi)\log(s-1) + \ell_{f_\chi}(s)$,
where $\ell_{f_\chi}(s)$ is a suitable logarithm of $f_\chi(s)$. Thus
$$
\log L(s,\chi) = n(\chi)\log(s-1) + O_\chi(1)
$$
for $s$ near $1$ to the right, and plugging this into the above displayed formula,
\begin{align}
\sum_{p \equiv a \bmod m} \frac{1}{p^s} & = \frac{1}{\varphi(m)}\sum_{\chi \bmod m} \overline{\chi}(a)(n(\chi)\log(s-1)+ O_\chi(1)) + O(1) \nonumber \\
& =\frac{1}{\varphi(m)}\left(\sum_{\chi} \overline{\chi}(a)n(\chi)\right)\log(s-1) + O_m(1).
\end{align}

To compute a Dirichlet density,
we want to divide both sides by $\sum_p 1/p^s$ for
$s$ near $1$ to the right. For such $s$,
$$
\log \zeta(s) = \sum_p \frac{1}{p^s} + O(1) = -\log(s-1) + O(1).
$$
Therefore $\sum_p 1/p^s \sim -\log(s-1)$ as $s \to 1^+$, so
dividing through by $\sum_p 1/p^s$ and letting $s \to 1^+$ gives us
\begin{equation}
\lim_{s \to 1^+}
\frac{\sum_{p \equiv a \bmod m} 1/p^s}{\sum_p 1/p^s} =
\frac{1}{\varphi(m)}\left(-\sum_{\chi} \overline{\chi}(a)n(\chi)\right),
\end{equation}
which expresses the Dirichlet density of $\{p \equiv a \bmod m\}$ in terms of the
orders of vanishing $n(\chi)$ as $\chi$ runs over Dirichlet characters mod $m$.

If (a) is true then $n(\chi) = 0$ for all nontrivial $\chi$, so the right side of the above limit calculation
is $(1/\varphi(m))(-n({\mathbf 1}_m)) = 1/\varphi(m)$, which is (b).

Conversely, if (b) is true then
$$
\sum_{\chi} \overline{\chi}(a)n(\chi) = -1
$$
for all $a \in (\mathbf Z/m\mathbf Z)^\times$ by our limit calculation. Why does this imply $n(\chi) = 0$ for nontrivial $\chi$?

Using complex vectors indexed by all the Dirichlet characters mod $m$, let
${\mathbf n}_m = (n(\chi))_\chi$ and
${\mathbf v}_a = (\chi(a))_\chi$ for each $a \in (\mathbf Z/m\mathbf Z)^\times$. The space of all complex vectors $\mathbf z = (z_\chi)_\chi$ has
dimension $\varphi(m)$ and it has the Hermitian inner product
$\langle \mathbf z, \mathbf w\rangle = \frac{1}{\varphi(m)}\sum_{\chi} z_\chi\overline{w_\chi}$ for which the vectors ${\mathbf v}_a$ are an orthonormal basis by the orthogonality relations for Dirichlet characters mod $m$. The above displayed formula
says $\langle {\mathbf n}_m,{\mathbf v}_a\rangle = -1/\varphi(m)$ for all $a$ in $(\mathbf Z/m\mathbf Z)^\times$, so
$$
{\mathbf n}_m = \sum_{a} \langle {\mathbf n}_m,{\mathbf v}_a\rangle{\mathbf v}_a = -\frac{1}{\varphi(m)}\sum_{a}{\mathbf v}_a.
$$
For each nontrivial $\chi \bmod m$, the $\chi$-component of
$\sum_{a} {\mathbf v}_a$ is $\sum_a \chi(a)$, which is $0$. So the $\chi$-component of ${\mathbf n}_m$, which is $n(\chi)$, is 0. That is (a).

QED Theorem 1. (I only realized after copying and pasting this that I had already copy and pasted it earlier as an answer to the MO question here.)

**Theorem 2**. For each $m \geq 1$, the following are equivalent.

a) For all Dirichlet characters $\chi \bmod m$, $L(s,\chi) \not= 0$ when ${\rm Re}(s) = 1$.

b) $\sum_{n \leq x} \chi(n)\Lambda(n) = o(x)$
for nontrivial Dirichlet characters $\chi \bmod m$
and $\sum_{n \leq x} \chi_{{\mathbf 1}_m}(n)\Lambda(n) \sim x$,

c) For all $a \in (\mathbf Z/m\mathbf Z)^\times$, $|\{p \leq x : p \equiv a \bmod m\}| \sim (1/\varphi(m))x/\log x$.

Comparing the proof of Theorem 2 below to the sketch in the answer by 2734364041, we will also be using a Tauberian theroem (to prove (b) implies (c)), but we will not need an explicit formula.

*Proof of Theorem 2*.

We will show (a) is equivalent to (b) and (b) is equivalent to (c).

First we show (a) implies (b).
Set $\psi_\chi(x) = \sum_{n \leq x} \chi(n)\Lambda(n)$ for all $\chi$, so
(b) says $\psi_\chi(x) = o(x)$ for nontrivial $\chi$ and $\psi_{{\mathbf 1}_m}(x) \sim x$.

For $\sigma > 1$, $-L'(s,\chi)/L(s,\chi) = \sum \chi(n)\Lambda(n)/n^s$, for all Dirichlet characters $\chi \bmod m$,
so $\psi_\chi(x)$ is a partial sum of coefficients of $-L'(s,\chi)/L(s,\chi)$.
Since $L(s,{\mathbf 1}_m) \not= 0$ on $\sigma = 1$ by (a), $-L'(s,{\mathbf 1}_m)/L(s,{\mathbf 1}_m)$ is holomorphic on $\sigma \geq 1$ except for a simple pole at $s = 1$ with residue 1 and it has nonnegative Dirichlet series coefficients with $\psi_{{\mathbf 1}_m}(x) = O(x)$. Therefore $\psi_{{\mathbf 1}_m}(x) \sim x$, which is part of (b), by Newman's Tauberian theorem. To get the rest of (b), namely $\psi_\chi(x) = o(x)$ for nontrivial $\chi$, we have $-L'(s,\chi)/L(s,\chi)$ being holomorphic on $\sigma \geq 1$ by (a) and its Dirichlet series coefficients satisfy $|\chi(n)\Lambda(n)| \leq {\mathbf 1}_m(n)\Lambda(n)$ for all $n$, so $\psi_\chi(x) = o(x)$ by a corollary of Newman's Tauberian theorem for $-L'(s,\chi)/L(s,\chi)$ using comparison Dirichlet series $-L'(s,{\mathbf 1}_m)/L(s,{\mathbf 1}_m)$ .

Thus (a) implies (b).

To show (b) implies (a), we will use the following fact. For a function $a(x)$ on $[1,\infty)$ that is bounded and Riemann integrable on $[1,T]$ for all $T \geq 1$, so $f(s) := \int_1^\infty (a(x)/x^s)dx/x$ is absolutely convergent on $\sigma > 1$, if $a(x) \to 0$ as $x \to \infty$ and $f$ extends to a meromorphic function on $\sigma = 1$ then $f$ in fact is holomorphic on $\sigma = 1$. (This is used to show the condition $\psi(x) \sim x$ implies $\zeta(s) \not= 0$ on $\sigma = 1$ by using $a(x) = \psi(x)/x - 1$.) Because of the integral representations
$$
-\frac{L'(s,\chi)}{sL(s,\chi)} = \int_1^\infty \frac{\psi_\chi(x)}{x} \frac{dx}{x^s}
$$
and
$$
-\frac{L'(s,{\mathbf 1}_m)}{sL(s,{\mathbf 1}_m)} - \frac{1}{s-1} = \int_1^\infty \left(\frac{\psi_{\mathbf 1}(x)}{x} -1\right)\frac{dx}{x^s},
$$
for ${\rm Re}(s) > 1$,
where $\chi$ is nontrivial in the first equation.
we can use the above fact when $a(x) = \psi_\chi(x)/x$ for nontrivial $\chi$ and
$a(x) = \psi_{{\mathbf 1}_m}(x)/x - 1$ to conclude that $L'(s,\chi)/L(s,\chi)$ is holomorphic on
$\sigma = 1$ for nontrivial $\chi$ and $L'(s,{\mathbf 1}_m)/L(s,{\mathbf 1}_m)$ is holomorphic on $\sigma = 1$ except for a simple pole at $s = 1$, so
$L(s,\chi)$ is nonvanishing on $\sigma = 1$ and
$L(s,{\mathbf 1}_m)$ is nonvanishing on $\sigma = 1$.
Thus (b) implies (a).

That (b) implies (c) follows from the above integral representation of $-L'(s,\chi)/L(s,\chi)$ for all nontrivial $\chi$ by a standard method to prove (c).

Our last step is showing (c) implies (b). Set $\pi(x;a \bmod m) = |\{p \leq x : p \equiv a \bmod m\}|$ when $(a,m) = 1$ and $\pi_\chi(x) = \sum_{p \leq x} \chi(p)$,
where $\chi$ is a Dirichlet character mod $m$.
Write $\chi$ as a linear combination of
delta-functions on $(\mathbf Z/m\mathbf Z)^\times$:
$\chi = \sum_{a \in (\mathbf Z/m\mathbf Z)^\times} \chi(a)\delta_a$. Then
\begin{align*}
\pi_\chi(x) & = \sum_{p \leq x} \chi(p) \\
& = \sum_{p \leq x} \sum_{a \in (\mathbf Z/m\mathbf Z)^\times} \chi(a)\delta_a(p) \\
& = \sum_{a \in (\mathbf Z/m\mathbf Z)^\times} \chi(a)\left(\sum_{p \leq x} \delta_a(p)\right) \\
& = \sum_{a \in (\mathbf Z/m\mathbf Z)^\times} \chi(a)\pi(x; a \bmod m),
\end{align*}
so
$$
\frac{\pi_\chi(x)}{x/\log x} = \sum_{a \in (\mathbf Z/m\mathbf Z)^\times} \chi(a)\frac{\pi(x;a \bmod m)}{x/\log x}.
$$
By (c), as $x \to \infty$ the right side tends to
$\sum_{a \in ({\mathbf Z}/m{\mathbf Z})^\times} \chi(a)/\varphi(m)$, which is 0 if $\chi$ is nontrivial. Therefore when $\chi$ is nontrivial we have $\pi_\chi(x) = o(x/\log x)$, which implies $\psi_\chi(x) = o(x)$ by
the same argument that $\pi(x) \sim x/\log x$ implies $\psi(x) \sim x$. To show (c) implies
$\psi_{{\mathbf 1}_m}(x) \sim x$, sum the relation in (c) over all $a$ in $(\mathbf Z/m\mathbf Z)^\times$ to
get $\pi(x) \sim x/\log x$, the Prime Number Theorem, which
is equivalent to
$\psi(x) \sim x$, so
$\psi_{{\mathbf 1}_m}(x) \sim x$ since $\psi_{{\mathbf 1}_m}(x) = \psi(x) + O_m(\log x)$.

QED Theorem 2.