It appears that for every odd prime $p$, the following congruence holds for Bernoulli numbers: $$ 2pB_{p-1}-pB_{2p-2}\equiv p-1\mod p^2\mathbb{Z}_{(p)}. $$ The weaker statement that $2pB_{p-1}-pB_{2p-2}\equiv -1\mod p\mathbb{Z}_{(p)}$ follows from the von Staudt-Claussen Theorem. I am aware of the Kummer congruences, but they don't seem to apply here as the indices of the Bernoulli numbers in question are divisible by $p-1$.
How does one prove this congruence modulo $p^2$?