Note $A$ is continuous:
$$\|A(x)\|=\left\|A\left(\sum_{n\in\Bbb N} x_n e_n\right)\right\| = \left\|\sum_n x_n y_n\right\|≤\sum_i|x_i|\,\|y_i\|≤\sum_i|x_i|=\|x\|$$
so its even contractive.
Let $y^*$ be in the unit ball of $E$, then there exists an $y_{n_1}$ with $\|y^*-y_{n_1}\|≤1/2$. It follows $2(y^*-y_{n_0})$ is in the unit ball of $E$ and you have a $y_{n_1}$ with $\|2(y^*-y_{n_0})-y_{n_1}\|≤1/2$, in other words $\|y^*-y_{n_0}-\frac{y_{n_1}}{2}\|≤2^{-2}$. Carry this out inductively to get a sequence $y_{n_k}$ so that
$$\left\|y^*-\sum_{l=0}^k2^{-l}y_{n_l}\right\|≤2^{-k-1}$$
So
$$A\left(\sum_{l=0}^\infty 2^{-l}e_{n_l}\right)=\sum_{l=0}^\infty 2^{-l}y_{n_l}=y^*$$
and the image of $A$ contains the unit ball. Since the unit ball is adsorbing it follows $A$ is surjective.
Remark: this proof can be adapted to show that if $A:X\to Y$ is a continuous map between Banach spaces so that $A(B_1(0)_X)$ is dense in $B_1(0)_Y$ that the map must be a surjection.