The implicit function theorem guarantees you a nice solution $x=x(u,v)$, $\ y=y(u,v)$ with $x(0,0)=0$, $\ y(0,0)=0$ if a certain "technical condition" is satisfied, but has nothing to say if this condition is not satisfied (which is the case here).
On the other hand we can go and solve the system for $x$ and $y$ explicitly. From the second equation we deduce $y=x u+\sin u-v^4$, and plugging this into the first equation gives $e^u+e^v+2=4\exp(x^2-x u-\sin u+v^4)$ or $x^2 - xu -\sin u +v^4-\log\Bigl(1+{e^u-1\over4}+{e^v-1\over4}\Bigr)=0\ .$ This is a quadratic equation for $x$ which you can solve explicitly, but it might have complex solutions for some $(u,v)$ near $(0,0)$. Maybe you don't allow this.
To see what happens we retain only the terms of order $\leq2$ in $u$ and $v$. We then have the equation $x^2-x u-{5\over4}u-{1\over4}v-{1\over32}(3u^2-2u v+3v^2)=0$ with the formal solutions $x={1\over2}\Bigl(u\pm\sqrt{5u+v+{\rm quadratic\ terms}}\Bigr)\ .$ Here the radicand is negative on one side of a quadric going through the origin of the $(u,v)$-plane. This implies that in the immediate neighborhood of $(0,0)$ there are points $(u,v)$ to which correspond no real $(x,y)$ solving the given system of equations.