 Solution

Yes. Your public and private exponents reveal $ed-1$, a nonzero multiple of $\lambda (N)$. The standard RSA private exponent reveals the factorization algorithm uses repeated squaring of random residues to obtain a nontrivial square root of one and hence factors $N$ with a greatest common divisor. Once $p$ and $q$ are known, compute $\varphi (N)$ and invert every other customer's public exponent to obtain that customer's private exponent.