Dense subset of continuous functions Let $C([0,1], \mathbb R)$ denote the space of real continuous functions at $[0,1]$, with the uniform norm. Is the set $H=\{ h:[0,1] \rightarrow \mathbb R : h(x)= \sum_{j=1}^n a_j e^{b_jx} , a_j,b_j \in \mathbb R, n \in \mathbb N \}$ dense in $C([0,1], \mathbb R)$?

Yes, it is. In place of Bernstein polynomials, we can use objects like:
$$ f_{j,k}(x)\triangleq C_{j,k}\cdot(e^x-1)^j\cdot(e-e^x)^k,\qquad \max_{x\in[0,1]}f_{j,k}(x)=1.$$
having a positive kernel in $H$. 
$f_{j,k}$ is concentrated around:
$$\operatorname{argmax} f_{j,k} = \log\frac{ej+k}{j+k}.$$
From this point on, we can follow the same lines of Bernstein's proof of the Weierstrass approximation theorem. Notice that we just took $b_j\in\mathbb{N}$.

As an alternative proof, consider that the sequence:
$$ g_n(x) = \tanh(nx)$$
pointwise converge towards $\operatorname{sign}(x)$ on $[-1,1]$, hence:
$$ G_n(x) = \frac{1}{n}\log\cosh(nx)$$
uniformly converges to $|x|$ on $[-1,1]$. Since there is a sequence of polynomials that uniformly converges to $\log\left(\frac{z^n+z^{-n}}{2}\right)$ for $z\in[e^{-1},e]$, we have that there exists a sequence of elements of $H$ that uniformly converges to $|x|$ on $[-1,1]$. Following now the lines of Lebesgue's proof of Weierstrass approximation theorem, we reach the same conclusion as before.

The shortest proof. Given $f\in C^0([0,1])$, consider $g:[1,e]\to\mathbb{R}$ given by $g(x)=f(\log x)$. By Weierstrass approximation theorem, there is a sequence of polynomials $\{g_n\}_{n\in\mathbb{N}}$ uniformly converging to $g$ on $[1,e]$. Then $\{g_n(e^x)\}_{n\in\mathbb{N}}$ is a sequence of elements of $H$ that uniformly converges to $g(e^x)=f$ on $[0,1]$.

A fourth proof, this time exploiting the fact that we are free to take $b_j\in\mathbb{R}$. Since, on $[0,1]$:
$$\left\|x-n\sinh\frac{x}{n}\right\|_{\infty}\leq \frac{\sinh(1)-1}{n^2}<\frac{3}{17\, n^2} $$
$$\left\|x-n\left(8\sinh\frac{x}{6n}-\sinh\frac{x}{3n}\right)\right\|_{\infty}<\frac{1}{38751\,n^4} $$
we can uniformly approximate $h(x)=x$ with elements of $H$, hence we can uniformly approximate any polynomial, hence any continuous function.
