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.
Update (2022): The prime number theorem
$\pi(x) \sim x/\log x$ is equivalent to two properties involving the Moebius function: $\sum_{n \leq x} \mu(n) = o(x)$ and
$\sum \mu(n)/n = 0$. These each have analogues in Theorem 2. See here.
