Is every true statement independent of $PA$ equivalent to some consistency statement?

The theory $PA + Con(PA)$ has the property you are asking for, this is the so called Friedman-Goldfarb-Harrington principle (see, e.g., Fifty years of self-reference in arithmetic, p. 366). Formally, for every $\Pi_1$ sentence $\pi$, there is a $\Pi_1$ sentence $\psi$ such that $PA + Con(PA) \vdash \pi \leftrightarrow Con(PA + \psi)$.

EDIT:

That being said, my hunch is that the answer must be "no" for plain $PA$ even though I can't produce a counterexample at the moment.

As observed by Will Sawin, we have that for every $\Pi_1$ sentence $\pi$, there is a $\Pi_1$ sentence $\psi$ such that $PA \vdash \pi \leftrightarrow Con(EA + \psi)$, which gives a positive answer to the OP's modified question.

My hunch is that $Con(EA + \psi)$ can not be replaced by $Con(PA + \psi)$ in the above.

$1$-consistency of $PA$ is a true $\Pi_3$ sentence which is not provable in $PA$+{all true $\Pi_1$ sentences} (see this article). Simple (iterated) consistency statements (as you mentioned above) are all (true) $\Pi_1$ sentences, so it is not equivalent to any $\Pi_1$ sentence.