Yes, the difference between $b_n$ and $a_n$ is always at least $1$. Let
$$f(n)=(1+1/n)^n,$$
so that $a_n=(n+1)^2f(n)$ and $b_n=(n+1)^2f(n+1).$ Then by the mean value theorem we have that
$$b_n-a_n=(n+1)^2(f(n+1)-f(n))=(n+1)^2f'(c)$$
for some $c\in (n,n+1)$. Next we calculate
$$f'(x)=f(x)\left(\log(1+1/x)-\frac{1}{x(1+1/x)}\right),$$
and after substituting in the power series expansions for the functions in the parenthesis we have
$$f'(x)=f(x)\left(\frac{1}{2x^2}-\frac{2}{3x^3}+\frac{3}{4x^4}-\cdots\right).$$
If $x>2$ then the terms in this series are decreasing in absolute value and
$$f'(x)\ge \frac{f(x)}{2x^2}\left(1-\frac{4}{3x}\right).$$
Keeping in mind that $f(x)$ is increasing, it is easy to check that this expression is greater than $1/x^2$ whenever $x>7$.
Substituting this back in the above equation gives $b_n-a_n>1$ for all $n>7$, and the rest is verified directly.