I seem to recall that the prime number theorem (PNT) is equivalent to the fact that the Riemann zeta function $\zeta(s)$ is non-zero on all of $\text{Re}(s) = 1$ (see https://math.stackexchange.com/questions/1379583/why-is-zeta1it-neq-0-equivalent-to-the-prime-number-theorem or https://math.stackexchange.com/questions/706934/riemann-zeta-function-non-vanishing-on-the-line-mathrmre-z-1), where I suppose "equivalent" means that PNT and the non-vanishing of $\zeta(s)$ along $\text{Re}(s)=1$ can be both be proven in a relatively short manner from the other.
Is it the case that Dirichlet's theorem on primes in arithmetic progressions is also equivalent to the fact that all the Dirichlet $L$-functions $L(s,\chi)$ are non-zero at $s=1$, in any sense similar to the equivalence described above? Perhaps another way of phrasing the question is: is it "possible" that $L(s,\chi)$ could equal 0 (for some non-trivial character $\chi$) but still Dirichlet's theorem could hold?
It has come to my attention that this has already been discussed here Analytic equivalents for primes in arithmetic progressions. Apologies all, and thanks for your comments.
Best Answer
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