Is Cramer-Shoup backdoorable?
A Cramer-Shoup public key has the form:


  $B = g_1^{b_1} g_2^{b_2}; \quad C = g_1^{c_1} g_2^{c_2}; \quad D = g_1^{d_1} g_2^{d_2}$


and an encryption of $m$ has the form:


  $g_1^r, ~ g_2^r, ~ m B^r, ~ (CD^\mu)^r$


where $\mu$ is a hash of the first 3 components.

CCA security proof breaks down:

In the CCA security proof, you observe that if you know the private key then you can generate the challenge ciphertext as a function of $g_1^r$ and $g_2^r$ (mirroring how the receiver would actually decrypt), which gives:


  $g_1^r, ~ g_2^r, ~ m [(g_1^r)^{b_1} (g_2^r)^{b_2}], ~ [ (g_1^r)^{c_1 + d_1 \mu} (g_2^r)^{c_2 + d_2 \mu} ]$


Then the first step of the proof is to replace $g_1^r, g_2^r$ with $g_1^{r_1}, g_2^{r_2}$ with $r_1, r_2$ random. This change is indistinguishable by DDH, but not if the discrete log of $g_2$ with respect to $g_1$ is known, as in your case. The rest of the proof follows from an information-theoretic argument.

Explicit CCA attack unclear:

Even though the proof seems to break down, I'm having trouble demonstrating an explicit attack on CCA security. You can definitely break the hybrids of  the CCA security proof without even consulting the decryption oracle. But we know the scheme remains CPA-secure (see below), so this it's possible that there's a problem only in the proof while the scheme remains CCA secure (though I doubt that).

CPA security maintained:

Even if you give the adversary everything but $b_1$ -- including the discrete log of $g_2$ with respect to $g_1$ -- you can still argue that the scheme is CPA secure. That's because we still have $r, b_1$ chosen uniformly, so $g_1^{r b_1}$ is pseudorandom given $g_1^r$, $g_1^{b_1}$. In particular, the third ciphertext component (the only one that involves the plaintext) is pseudorandom.