Simultaneity problem: use of Instrumental Variables and test
To clarify notation, and treating summarily the other variables we have a system

$$y = a_0 +a_1x + \mathbf w_1'\gamma_1 + u_1 \\
x = b_0 + b_1y + \mathbf w_2'\gamma_2 + u_2$$

If $b_1 \neq 0$, then solving for $x$ we obtain

$$ x = b_0 + b_1(a_0 +a_1x + \mathbf w_1'\gamma_1+ u_1) + \mathbf w_2'\gamma_2 + u_2 \\
\implies x = c + \mathbf w_1'\frac {\gamma_1}{1-a_1b_1}+\frac {b_1}{1-a_1b_1}u_1 + \mathbf w_2'\frac {\gamma_2}{1-a_1b_1}+\frac {1}{1-a_1b_1}u_2$$

and  $x$ is correlated with $u_1$ in the first equation, and so it is endogenous (we would obtain an analogous result if we solved for $y$ -it would be endogenous in the second equation). Historically, this is the "original" endogeneity problem, the one realized back in the '20s when looking at market equilibrium, where the assumption of market-clearing created the endogeneity. Namely it is not the usual "omitted variables" endogeneity problem, but the form of endogeneity due to "simultaneity".  

So endogeneity of $x$ hinges on $b_1\neq 0$. But running OLS on the second equation and testing whether $b_1=0$ is not valid, because, as said, $y$ is endogenous in the second equation, so the OLS estimation is unreliable, being inconsistent.
This is a bit subtle. We want to test $b_1 =0$, so under the null hypothesis $y$ does not belong in the second equation. But inference will use $y$ itself. This means that if the null is rejected, the rejection is not reliable, because if the null is not true, then $b_1 \neq 0$ and automatically the estimator, t-tests etc become inconsistent.

The question asks to test for endogeneity in the first equation, so the evident thing to do is to search for an instrument, or instruments, for $x$ and then, for example, apply the Hausman test in the first equation.

Assume that we applied the Hausman test in the second equation, as the OP suggested. This would mean searching for instruments for $y$. Here if the null is not rejected, it would mean that $y$ is not correlated with $u_2$. But from this it does not necessarily follow that $b_1\neq 0$, and so it does not guarantee that $x$ is uncorrelated with $u_1$. If the null is rejected, it means that $y$ is correlated with $u_2$. But this could happen for any other reasons, not necessarily due to the simultaneity.
So the endogeneity tests should be conducted on the first equation. 

Note that in reality, the Hausman test  in IV estimation has as its null hypothesis that both the OLS and the IVE estimator are consistent. If the null is rejected, it only tells us that at least one of the two is inconsistent. It cannot tell us which one, and it cannot guarantee that not both are inconsistent: The Hausman test is more informative and specific in its conclusions when the null is not rejected, rather than when the null is rejected. (This is a point usually not clearly stressed in the literature, although side remarks and footnotes do exist to that effect).