Given two different safe primes $p_1$, $p_2$ we construct the subgroups $G_1$ of prime order $q_1 = \frac{p_1-1}{2}$ and $G_2$ of prime order $q_2 = \frac{p_2-1}{2}$. Let $g_1$ be a generator of $G_1$ and $g_2$ be a generator of $G_2$. We assume that the discrete logorithm is hard in both groups.
Alice randomly selects some secret value $ s \in \mathbb{Z}_{min\{q_1, q_2\}} $. Alice calculates $ x_1 := g_1^s \in G_1 $ and $ x_2 := g_2^s \in G_2 $ and a cryptographic hash $ H(s) $.
Alice sends $x_1$, $x_2$ and $H(s)$ to Bob via a secure channel.
Is there some pratical way that Bob can be sure that Alice indeed has sent values $ x_1 = g_1^a $ and $ x_2 = g_2^b $ such that $ a = b $? If not, can Alice provide additional data without revealing $s$ to convince Bob?
This question is somewhat related to "How can we prove that two discrete logarithms are equal?". However in our case two generators from two different groups are used instead of two generators from one group.
EDIT: Regarding the comments from poncho and Yehuda Lindell I further clarify what I mean by "discrete logarithm equality for independent groups":
I would like to use the following definition for $dlog_{g}$:
$ {dlog}_g(g^e) := $ smallest $ x \mid x \geqslant 0, g^x \equiv g^e \pmod p $
Given $ s_1 = dlog_{g_1}(g_1^s) $ and $ s_2 = dlog_{g_2}(g_2^s) $
then $ dlog_{g_1}(g_1^s) \ $ equals $ \ dlog_{g_2}(g_2^s) \ $ iff $ \ s_1 = s_2 $