As a hint, your first step is incorrect. Once you've tried building each side independently, you will not have a way to ensure that they have the same number of 1s.
Instead, try building the string from the outside inward. If you were going to put 0s and 1s into each half of the string, how would you ensure that you did so in a way that always ensured that the number of 0s and 1s was the same?
Hope this helps!
While it’s true that you will eventually eliminate the production $A\to\lambda$, you can’t just throw it away: by doing so, you’ve changed the language generated by the grammar. The original grammar allows the derivation $$S\Rightarrow aAb\Rightarrow ab\;,$$ while your new one cannot generate the word $ab$ at all. To see this, note that every one of your productions increases the length of the derived string, and the only terminal production is $A\to bb$; it takes at least one other step just to get an $A$, so your grammar cannot possibly produce any word of length less than $3$. (In fact the shortest possible length is $4$, for $abbb$.)
The idea is to adjust all productions that have $A$ on the righthand side in a way that compensates for losing the production $A\to\lambda$. If we have a production $X\to vAw$, where $v$ and $w$ are any strings of terminal and/or non-terminal symbols, the production $A\to\lambda$ allowed the derivation $X\Rightarrow vAw\Rightarrow vw$. If we throw out the production $A\to\lambda$, we lose this possibility, so to compensate we add the production $X\to vw$; this permits the derivation $X\Rightarrow vw$, which has the same effect as the original derivation $X\Rightarrow vAw\Rightarrow vw$ that is no longer available.
In particular, this means that we must add the production $S\to ab$ alongside the existing $S\to aAb$ and the production $B\to ba$ alongside the existing $B\to bAa$.
Taking care of $B\to AA$ is a little trickier, but it’s still not bad if you think in terms of compensating for loss of $A\to\lambda$. In the original grammar we have derivations
$$\begin{align*}
&B\Rightarrow AA\Rightarrow A\lambda=A\text{ and}\\
&B\Rightarrow AA\Rightarrow\lambda A=A
\end{align*}$$
that are no longer available when we throw out $A\to\lambda$, so we have to add $B\to A$. But we also have
$$B\Rightarrow AA\Rightarrow^*A\Rightarrow\lambda\;,$$
in which we apply the $\lambda$ production to both $A$’s, so we also need to add $B\to\lambda$. At this point we have the following productions:
$$\begin{align*}
&S\to aAb\mid ab\mid BBa\\
&A\to bb\\
&B\to AA\mid A\mid\lambda\mid bAa\mid ba\;.
\end{align*}$$
We got rid of $A\to\lambda$, but at the cost of introducing a new $\lambda$ production, $B\to\lambda$. That’s okay: we can repeat the process to get rid of $B\to\lambda$. The only production with $B$ on the righthand side is $S\to BBa$. The derivations that we lose by throwing away $B\to\lambda$ are $$S\Rightarrow BBa\Rightarrow Ba$$ and $$S\Rightarrow BBa\Rightarrow Ba\Rightarrow a\;,$$ so we can compensate for the loss of $B\to\lambda$ by adding productions $S\to Ba$ and $S\to a$:
$$\begin{align*}
&S\to aAb\mid ab\mid BBa\mid Ba\mid a\\
&A\to bb\\
&B\to AA\mid A\mid bAa\mid ba\;.
\end{align*}$$
Best Answer
I think your solution is incomplete, how do you know that $w_1$ is in $L(G)$. Consider $w=aaabbb$,then $w1=aabbb$ which clearly is not in the language.
I will put my answer below, if you want to think more on it you can stop here.
Now on, by the property I mean :"every prefix of $w$ has at least as much $a$s as $b$s".
The proof must consist of two parts: 1:proving every string produced by the grammar has the aforementioned property, 2:proving every string with the property is produced by the grammar. This is the way I would prove it.
For part one, suppose the property holds for all strings produced by the grammar like $u$ where $|u|<n$. Now suppose $|w|=n>0$, then the first production in derivation of $w$ must be $S\to aS$ or $S\to aSbS$. In either case $S$ symbols will be replaced with strings in $L(G)$ with length less than $n$, and we know in such strings the property holds. Then it is easy to show that it must also hold in $w$.
For part two, suppose every string with the property with length less than $n$ is produced by the grammar, suppose $w$ with length $n>0$ has the property. Then $w=av,|v|<n$, now if $v$ also has the property then $w$ can be made with a sequence of derivations starting with $S\to aS$. If $v$ does not have the property then $v$ must have at least one $b$, consider the last $b$ from left to right in $w$,then $w=axby$. Note that the $b$ is the last $b$ and $y$ can be $\epsilon$. Since $w$ has the property, then $axb$ has the property, then $x$ hast the property and $|x|<n$. Also, as there are no $b$s in $y$, then it has the property too, so $w$ can be made with a sequence of derivations starting with $S\to aSbS$, the first $S$ makes $x$ and the second one makes $y$.
P.S: You should add base case for the inductions, I didn't mention them.