If [tex]f(z)=f(x+iy)=u(x,y)+iv(x,y)[/tex] is analytic, then the Cauchy-Riemann conditions hold:
[tex]\begin{cases}\dfrac{\partial u}{\partial x}=\dfrac{\partial v}{\partial y}\\\\\dfrac{\partial u}{\partial y}=-\dfrac{\partial v}{\partial x}\end{cases}[/tex]
I'll denote these partial derivatives by [tex]u_x,v_y,u_y,v_x[/tex].
You have
[tex]\begin{cases}(u-v)_x=u_x-v_x=e^x(\cos y-\sin y)&(1)\\(u-v)_y=u_y-v_y=e^x(-\sin y-\cos y)&(2)\end{cases}[/tex]
Adding and subtracting [tex](1)[/tex] and [tex](2)[/tex], respectively, gives
[tex]\begin{cases}(u_x-v_y)+(u_y-v_x)=-2e^x\sin y&(1)+(2)\\(u_x+v_y)-(u_y+v_x)=2e^x\cos y&(1)-(2)\end{cases}[/tex]
Because [tex]f(z)[/tex] is analytic, this system reduces to
[tex]\begin{cases}2u_y=-2e^x\sin y&(3)\\2u_x=2e^x\cos y&(4)\end{cases}\iff\begin{cases}u_y=-e^x\sin y&(3)\\u_x=e^x\cos y&(4)\end{cases}[/tex]
Integrating [tex](3)[/tex] with respect to [tex]y[/tex] gives
[tex]\displaystyle\int u_y\,\mathrm dy=\int-e^x\sin y\,\mathrm dy[/tex]
[tex]u=e^x\cos y+g(x)[/tex]
Differentiating with respect to [tex]x[/tex] gives
[tex]u_x=e^x\cos y+g'(x)=e^x\cos y\implies g'(x)=0\implies g(x)=C_1[/tex]
where [tex]C_1[/tex] is an arbitrary constant.
On the other hand, you have
[tex]\begin{cases}-v_x=-e^x\sin y&(5)\\v_y=e^x\cos y&(6)\end{cases}[/tex]
(Note that [tex](5)=(3)[/tex] and [tex](6)=(4)[/tex] by the C-R conditions.) Integrating [tex](5)[/tex] with respect to [tex]x[/tex] gives
[tex]\displaystyle\int-v_x\,\mathrm dx=\int-e^x\sin y\,\mathrm dx[/tex]
[tex]v=e^x\sin y+h(y)[/tex]
Differentiating with respect to [tex]y[/tex] gives
[tex]v_y=e^x\cos y+h'(y)=e^x\cos y\implies h'(y)=0\implies h(y)=C_2[/tex]
It follows that
[tex]f(x+iy)=(e^x\cos y+C_1)+i(e^x\sin y+C_2)[/tex]
[tex]f(x+iy)=e^x(\cos y+i\sin y)+(C_1+iC_2)[/tex]
[tex]f(z)=e^{iz}+C_3[/tex]
where the last equality is an invocation of Euler's formula.
