Let us show that $r(x):=\frac{x-1}{N^{-1}(\Phi(x))}$ is increasing in $x\ge1$. The condition $r'(x)>0$ can be rewritten as $y n(y)>k(x-1)n(x-1)$, where $y:=y(x):=N^{-1}(\Phi(x))$ and $k:=\frac{\sqrt{2\pi}}{2+\sqrt{2\pi}}\in(0,1)$. You already know that $N(x-1)<\Phi(x)$, which is equivalent to $y>x-1$. So, it remains to show that $n(y)>k n(x-1)$, which can be rewritten as $y<z(x):=\sqrt{(x-1)^2-2\ln k}$ and then as $\Phi(x)<N(z(x))$ or as $1-k+k N(x-1)<N(z(x))$. Differentiating both sides of the latter inequality, we see that it is enough to show that $k n(x-1)>n(z(x))z'(x)$, which follows because $k n(x-1)=n(z(x))$ and $z'(x)=(x-1)/z(x)<1$.
To show that $r(x)=\frac{x-1}y\to1$, recall the well-known relation $1-N(x)\sim n(x)/x$, which implies $1-N(x)=\exp\{-\frac12\,x^2\,(1+o(1))\}$; everything in this paragraph is taken as $x\to\infty$. The definition $y:=N^{-1}(\Phi(x))$ can be rewritten as $N(y)=\Phi(x)$ and then as $1-N(y)=k(1-\Phi(x))$, which shows that $y\to\infty$ and $\exp\{-\frac12\,y^2\,(1+o(1))\}=\exp\{-\frac12\,x^2\,(1+o(1))\}$. So, $y\sim x$; that is, $r(x)=\frac{x-1}y\to1$.