First, the following grammar produces the language $L_0$ containg words where $n(b) \neq n(a)$
$$S \rightarrow U \;|\; V $$
$$U \rightarrow TbT \;|\; TbU$$
$$V \rightarrow TaT \;|\; TaV$$
$$T \rightarrow bTaT \;|\; aTbT \;|\; \epsilon $$
It follows your idea of making two cases, and $T$ is the one giving words with equality and is the only non-obvious (but classic) one.
We have a map from $L$ to $L_0$ by transforming every $b$ into an $bb$.
Let's try to pull this grammar back through this map.
In order to do this, we need a grammar for the image of this map :
We introduce new symbols $U_b, V_b, T_b$ where $b U_b$ is intended to decompose into the words that $U$ can decompose into but beginning with an $b$.
We do the same and have new symbols $U^b, V^b, T^b$ where $U^b b$ decomposes into the words $U$ can decompose into but ending with an $b$.
We get (you have to be careful when a $T$ appears because it can be empty) :
$$ S \rightarrow U \;|\; V $$
$$ U \rightarrow (T^b b)bT \;|\; Tb(bT_b) \;|\; (T^b b)bU \;|\; Tb(bU_b) $$
$$ U_b \rightarrow (T_b^b b)bT \;|\; T_b b(bT_b) \;|\; T
\;|\; (T_b^b b)bU \;|\; T_bb(bU_b) \;|\; U $$
$$ V \rightarrow TaT \;|\; TaV $$
$$ T \rightarrow b(bT_b)aT \;|\; a(T^bb)bT \;|\; aTb(bT_b) \;|\; \epsilon$$
$$ T_b \rightarrow TaT$$
$$ T^b \rightarrow b(bT_b)aT^b \;|\; a(T^bb)bT^b \;|\; aTb(bT_b^b) \;|\; aT$$
$$ T_b^b \rightarrow TaT^b$$
which describes the words of $L_0$ where every $b$ appears in a pair $bb$.
Now we can pull back and simplify the two symbols having only one rule and get this :
$$ S \rightarrow U \;|\; V $$
$$ U \rightarrow T^b bT \;|\; TbTaT \;|\; T^b bU \;|\; TbU_b $$
$$ U_b \rightarrow TaT^b bT \;|\; TaTbTaT \;|\; T
\;|\; TaT^b bU \;|\; TaTbU_b \;|\; U $$
$$ V \rightarrow TaT \;|\; TaV $$
$$ T \rightarrow bTaTaT \;|\; aT^bbT \;|\; aTbTaT \;|\; \epsilon$$
$$ T^b \rightarrow bTaTaT^b \;|\; aT^bbT^b \;|\; aTbTaT^b \;|\; aT$$
Which should work, if I didn't mess up. Note that this method can be generalized to pulling back any grammar through any such map.