Formalizing composition of sequences

You can use the basic fact that for any sequence $c_n \stackrel{n\to\infty}{\longrightarrow}0$ $(c_n \neq 0)$ you have

$$\left(1+c_n\right)^{\frac 1{c_n}}\stackrel{n\to\infty}{\longrightarrow}e$$

So with $c_n =\frac{b_n}n$ and $b_n \stackrel{n\to\infty}{\longrightarrow}b$, you get

\begin{eqnarray*} \left(1+c_n\right)^n & = & \left(\left(1+c_n\right)^\frac 1{c_n}\right)^{c_n\cdot n}\\ & = & \left(\left(1+c_n\right)^\frac 1{c_n}\right)^{b_n} \\ & \stackrel{n\to\infty}{\longrightarrow} & e^b \end{eqnarray*}

Here the continuity of $f(x,y) = x^y$ in both variables $x$ and $y$ is used which can easily be established considering,for example, first the logarithm of this function.

Hence, you have

$$\lim_{n\to\infty} \left(1+\frac{b_n}n\right)^n=e^b = \lim_{n\to\infty} \left(1+\frac{b}n\right)^n$$

The desired statement is true. It follows from taking logarithms and writing

$$\begin{eqnarray*} \log \left( 1 + \frac{b_n}{n} \right)^n &=& n \log \left( 1 + \frac{b_n}{n} \right) \\ &=& n \left( \frac{b_n}{n} + O \left( \frac{b_n^2}{n^2} \right) \right) \\ &=& b_n + O \left( \frac{b_n^2}{n} \right). \end{eqnarray*}$$

Taking the limit as $n \to \infty$ gives that the limit of the logarithms is $\lim_{n \to \infty} \left( b_n + O \left( \frac{b_n^2}{n} \right) \right) = b$ (using that $b_n$ is convergent and hence bounded, so the $O \left( \frac{b_n^2}{n} \right)$ term vanishes in the limit).

Depending on how you've defined exponentials and logarithms and proven their properties this argument may be circular, since it depends on the first two terms of the Taylor series of the logarithm, and one of several ways to compute the Taylor series of the logarithm is to use $\log xy = \log x + \log y$ which is more or less equivalent to $e^{x+y} = e^x e^y$.

There are alternatives for proving that $e^{x + y} = e^x e^y$ which avoid having to prove this; for example, taking for granted that $\frac{d}{dx} e^x = e^x$ and $e^0 = 1$ (which together uniquely determine $e^x$ by the existence and uniqueness theorems for ODEs), we have first

$$\frac{d}{dx} (e^x e^{-x}) = e^x e^{-x} - e^x e^{-x} = 0$$

so $e^x e^{-x}$ is a constant, and plugging in $x = 0$ gives $e^x e^{-x} = 1$. Next,

$$\frac{d}{dx} \left( e^{x+y} e^{-x} e^{-y} \right) = e^{x+y} e^{-x} e^{-y} - e^{x+y} e^{-x} e^{-y} = 0$$

so $e^{x+y} e^{-x} e^{-y}$ is a constant in $x$, and plugging in $x = 0$ gives $e^y e^{-y} = 1$ and hence $e^{x+y} = e^x e^y$.

Another approach: Note that
$$\left(1+\frac{x+y+\frac{xy}{n}}{n}\right)^n = \left(1+\frac x n\right)^n\left(1+\frac y n\right)^n.$$

Addendum

We have that \begin{align*} \left(\frac{1+\frac{x+y+\frac{xy}{n}}{n}}{1+\frac{x+y}{n}}\right)^n &= \left( \frac{\left(1+\frac x n\right)\left(1+\frac y n\right)}{1+\frac{x+y}{n}}\right)^n \longrightarrow \frac{e^x e^y}{e^{x+y}} \quad(n\to\infty). \end{align*} It remains to show that this limit is unity. See this answer, for example.