Given the function $y=x^{2}+2x+3=(x+1)^{2}+2$ with $x\geqslant 0$,

We observe that the function is a quadratic function in the form of $(x+1)^{2}+2$. Since $x\geqslant 0$, the function is always increasing for $x\in[0,+\infty)$.

To find the minimum value of the function, we can set $x=0$ because the function is increasing. Thus, the minimum value is $3$.

Moreover, the function does not have a maximum value since it keeps increasing as $x$ grows.

Therefore, the range of the function is $[3,+\infty)$.

Hence, the correct answer is: $\boxed{D}$.

This solution is obtained by utilizing the properties of quadratic functions and analyzing the given function's behavior based on its form. This problem primarily tests the application of quadratic function properties and can be considered as a basic question.