In a additive group is it hard to calculate $bg$ given $ag, g, abg$ The ECDH problem defined that given $g,ag,bg$ it is difficult to calculate $abg$. But it is also difficult to calculate $bg$ given $ag,g,abg$. where $g$ is generator and a,b are elements of group.
Yes, it is equally as difficult; if we assign:

$$g' = ag$$
$$a' = a^{-1}$$
$$b' = b$$

Then the restatement of your problem is: given $g' = ag$, $a'g' = g$ and $b'g' = abg$, compute $a'b'g' = bg$, which is exactly the ECDH problem.

Now, this assumes that $a$ has an inverse; this is not a problem if the curve order is a prime, and is easy to work around if the curve order is a prime times a small cofactor.