Theorem: Fix a positive integer $m$. The following two conditions are equivalent:
(1) $L(1,\chi) \not= 0$ for all nontrivial Dirichlet characters $\chi \bmod m$.
(2) 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. When $m$ is $1$ or $2$, the condition (1) is vacuously true (there are no nontrivial Dirichlet characters mod $m$) and condition (2) is obvious, so from now on we can let $m \geq 3$.
The usual proof of Dirichlet's theorem shows (1) implies (2). It remains to show (2) implies (1).
For each Dirichlet character $\chi \bmod m$, $L(s,\chi)$ is analytic for ${\rm Re}(s) > 0$ except that $L(s,\mathbf 1_m)$ has a simple pole at $s = 1$, where $\mathbf 1_m$ denotes the trivial character mod $m$ (we have $L(s,\mathbf 1_m) = \zeta(s)\prod_{p \mid m} (1-1/p^s)$ and $\zeta(s)$ is analytic on ${\rm Re}(s) > 0$ except for a simple pole at $s = 1$).
Set
$$
r_\chi := {\rm ord}_{s=1}(L(s,\chi))
$$
so $r_{\mathbf 1_m} = -1$ (the simple pole at 1) and $r_\chi \geq 0$ for all nontrivial $\chi$. Condition (1) is equivalent to $r_\chi = 0$ for all nontrivial $\chi$, so we want to show condition (2) implies the numbers $r_\chi$ vanish for all nontrivial $\chi$.
For a Dirichlet character $\chi \bmod m$ and ${\rm Re}(s) > 1$, define
$$
\log L(s,\chi) := \sum_{p,k} \frac{\chi(p^k)}{kp^{ks}} =
\sum_p \frac{\chi(p)}{p^s} + \sum_{p,k\geq 2} \frac{\chi(p^k)}{kp^{ks}}.
$$
This is a logarithm of $L(s,\chi)$ (meaning the exponential of that series is $L(s,\chi)$. The second series on the right is absolutely convergent for ${\rm Re}(s) > 1/2$, with absolute value bounded by $\sum_{p,k \geq 2} 1/(kp^{k\sigma})$, so for $s > 1$ we can say
$$
\log L(s,\chi) = \sum_p \frac{\chi(p)}{p^s} + O(1),
$$
where the $O$-constant is $\sum_{p,k \geq 2} 1/(kp^k)$. In the usual proof of Dirichlet's theorem, for each $a \in (\mathbf Z/m\mathbf Z)^\times$ and $s > 1$ we write
$$
\sum_{p \equiv a \bmod m} \frac{1}{p^s} =
\frac{1}{\varphi(m)}\sum_{\chi \bmod m} \overline{\chi}(a)\left(\sum_p \frac{\chi(p)}{p^s}\right),
$$
so
$$
\sum_{p \equiv a \bmod m} \frac{1}{p^s} =
\frac{1}{\varphi(m)}\sum_{\chi \bmod m} \overline{\chi}(a)\log L(s,\chi) + O(1),
$$
where the $O$-constant on the right is an overall term (outside the sum).
Now let's bring in the order of vanishing $r_\chi$. For all $s$ near $1$, $L(s,\chi) = (s-1)^{r_\chi}f_\chi(s)$ where $f_\chi(s)$ is a holomorphic function in a neighborhood of $s = 1$ (in fact it is holomorphic on ${\rm Re}(s) > 0$, or even $\mathbf C$) and $f_\chi(1) \not= 0$. Therefore $f_\chi(s)$ has a logarithm around $s = 1$ (well-defined up to adding an integer multiple of $2\pi$), so for $s > 1$
$$
\log L(s,\chi) = r_\chi\log(s-1) + \ell_{f_\chi}(s)
$$
where $\ell_{f_\chi}(s)$ is a suitable logarithm of $f_\chi(s)$. Thus
$\log L(s,\chi) = r_\chi\log(s-1) + O(1)$ for $s$ near $1$ to the right, and plugging this into the formula for
$\sum_{p \equiv a \bmod m} 1/p^s$ we can say
$$
\sum_{p \equiv a \bmod m} \frac{1}{p^s} =
\frac{1}{\varphi(m)}\sum_{\chi \bmod m} \overline{\chi}(a)(r_\chi\log(s-1)) + O(1).
$$
Let's extract the term for the trivial character mod $m$: since
$r_{\mathbf 1_m} = -1$,
$$
\sum_{p \equiv a \bmod m} \frac{1}{p^s} = -\frac{1}{\varphi(m)}\log(s-1) +
\frac{1}{\varphi(m)}\left(\sum_{\chi \not= \mathbf 1_m} \overline{\chi}(a)r_\chi\right)\log(s-1) + O(1).
$$
In order to bring in a Dirichlet density,
we want to divide both sides by $\sum_p 1/p^s$ for
$s$ near $1$ to the right. For such $s$,
$$
\log \zeta(s) = -\log(s-1) + O(1)
$$
from the simple pole of $\zeta(s)$ at $s = 1$ and
$$
\log \zeta(s) = \sum_p \frac{1}{p^s} + O(1)
$$
from the Euler product for $\zeta(s)$ when $s > 1$. Therefore $\sum_p 1/p^s = -\log(s-1) + O(1)$ as $s \to 1^+$, so $-\log(s-1) \sim \sum_p 1/p^s$ as $s \to 1^+$.
Dividing through the last (big) formula above for $\sum_{p \equiv a \bmod m} 1/p^s$ by $\sum_p 1/p^s$ and letting $s \to 1^+$, we get
$$
\frac{\sum_{p \equiv a \bmod m} 1/p^s}{\sum_p 1/p^s} \to \frac{1}{\varphi(m)} -
\frac{1}{\varphi(m)}\left(\sum_{\chi \not= \mathbf 1_m} \overline{\chi}(a)r_\chi\right)
$$
as $s \to 1^+$. So we have shown, without assuming condition (1) in the theorem, that for all $a \in (\mathbf Z/m\mathbf Z)^\times$ the set of primes $\{p \equiv a \bmod m\}$ has Dirichlet density
$$
\frac{1}{\varphi(m)}\left(1 - \sum_{\chi \not= \mathbf 1_m} \overline{\chi}(a)r_\chi\right).
$$
Finally it is time to assume condition (2) in theorem, which implies
$$
\sum_{\chi \not= \mathbf 1_m} \overline{\chi}(a)r_\chi = 0
$$
for all $a \in (\mathbf Z/m\mathbf Z)^\times$.
When $\chi = \mathbf 1_m$, $\overline{\chi}(a)r_\chi = 1(-1) = -1$, so condition (2) implies
$$
\sum_{\chi} \overline{\chi}(a)r_\chi = -1
$$
for all $a \in (\mathbf Z/m\mathbf Z)^\times$,
where the sum runs over all Dirichlet characters mod $m$ (including the trivial character). We want to show the above equation, for all $a$, implies $r_\chi = 0$ for all nontrivial $\chi \bmod m$.
Using vectors indexed by the Dirichlet characters mod $m$, let
$\mathbf r_m = (r_\chi)_\chi$ 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 a 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 usual orthogonality of Dirichlet characters mod $m$. The equation
$\sum_\chi \overline{\chi}(a)r_\chi = -1$ above says
$\langle \mathbf r_m,\mathbf v_a\rangle = -1/\varphi(m)$ for
all $a$ in $(\mathbf Z/m\mathbf Z)^\times$, so
$$
\mathbf r_m = \sum_{a} \langle \mathbf r_m,\mathbf v_a\rangle\mathbf v_a =
-\frac{1}{\varphi(m)}\sum_{a}\mathbf v_a.
$$
For nontrivial Dirichlet characters $\chi \bmod m$, the $\chi$-component of $\sum_{a} \mathbf v_a$ is
$\sum_a \chi(a)$, which is $0$ (the $\mathbf 1_m$-component is $\varphi(m)$, but that's irrelevant). Since
$\mathbf r_m$ has $\chi$-component $r_\chi := {\rm ord}_{s=1}L(s,\chi)$, we have
$r_\chi = 0$ for all nontrivial $\chi$, so
$L(1,\chi) \not= 0$ for all nontrivial $\chi$. QED
