Let $p$ be the characteristic of $\mathfrak{p}$.
If $p\nmid N$, then the map is an isomorphism, since both groups have order $N^2$ and the kernel is in the formal group, which has no prime-to-$p$ torsion. [1, Prop VII.3.1] So by the Chinese remainder theorem, we're reduced to the case that $N=p^k$. If $\tilde E$ is supersingular, then $\tilde E[p^k]=0$ for all $k$, so there's nothing to prove. [1, V Sec 4] The remaining case is $\tilde E$ ordinary, in which case $\tilde E[p^k]\cong\mathbb{Z}/p\mathbb{Z}$. There are a number of ways to do this case. One is to again use the formal group and note that when the curve has ordinary reduction, then it's formal group has height 1, so the kernel contains one copy of $\mathbb{Z}/p\mathbb{Z}$, which means that the image must map onto $\tilde E[p]\cong\mathbb{Z}/p\mathbb{Z}$. This isn't explicitly stated in [1], but can be pieced together from material in chapters IV, V, and VII.
[1] The Arithmetic of Elliptic Curves, J.H. Silverman, Springer