My claim is that it can. Normally, you divide the coefficient by the standard error and compare to a t distribution: b/se ~ t

Instead, you can just compare the coefficient to a scaled t distribution: b ~ t(se)

Thus, for your test, you compare the coefficient to a calculable distribution under the null and it should be exactly equivalent to the normal setup. Therefore, the coefficient *can* operate as a test statistic. Correct logic?