Writing $f(z)=\sum_{n=0}^\infty c_n z^n$, we see that $$u=\frac12(f+\bar f) = \frac12\sum_{n=0}^\infty (c_n z^n+\bar c_n\bar z^n)\tag1$$
The $n$th term in the series in (1) is a homogeneous polynomial in $x,y$ of degree $n$. Therefore, there is no cancellation of monomials $x^ky^m$ between the terms with different $n$.
Conclusion: for $d\ge 1$, $u$ is a harmonic polynomial of degree $d$ if and only if $f$ is a complex polynomial of degree $d$.
(The case $d=0$ is a bit special, since zero constant and nonzero constants have different degrees.)
The coefficient of $x^n$ in (1) is $\operatorname{Re}c_n$ while the coefficient of $x^{n-1}y$ is $i\operatorname{Im}c_n$. This gives you $c_n$ for every $n\ge 1$. (Of course, for $n=0$ you don't know $\operatorname{Im}c_0$ for the lack of $x^{n-1}y$.) Having $c_n$, you can write down
$$v=\frac12(f-\bar f) = \frac12\sum_{n=0}^\infty (c_n z^n-\bar c_n\bar z^n)\tag2$$
and expand it into $x$ and $y$ monomials if you wish.
(Personally, I never find much need to expand harmonic functions into $x^ky^m$ monomials; the expansion into powers of $z$ and of $\bar z$ is more efficient because it encodes the harmonicity of the function.)
Let $V(x, y)$ be the vector field on $\Bbb R^2$ given by
$\vec V(x, y) = \begin{pmatrix} -e^x\cosh y \\ e^x \sinh y \end{pmatrix}; \tag 1$
note that the components of $\vec V(x, y) = (V_x(x, y), V_y(x, y))$ are the two functions
$V_x(x, y) = -e^x\cosh y \tag 2$
and
$V_y(x, y) = e^x \sinh y \tag 3$
which would be the $x$- and $y$-components of $\nabla v(x, y)$, the gradient of $v(x, y)$, provided such a function $v(x, y)$ were to exist. So we seek to know if $\vec V(x, y)$ is the gradient of some function. This may be quite easily resolved in the present case by calulating $\nabla \times \vec V(x, y)$, since we know that a vector field is a gradient if and only if its curl vanishes. We have
$\nabla \times \vec V(x, y) = \det \left (\begin{bmatrix} \mathbf i & \mathbf j & \mathbf k \\ \partial_x & \partial_y & \partial_z \\ V_x(x, y) & V_y(x, y) & 0 \end{bmatrix} \right )$
$= (\partial_y(0) - \partial_z V_y) \mathbf i + (\partial_z V_x - \partial_x(0))\mathbf j + (\partial_x V_y - \partial_y V_x)\mathbf k = (\partial_x V_y - \partial_y V_x)\mathbf k ; \tag 4$
now,
$\partial_x V_y(x, y) - \partial_y V_x(x, y) = e^x \sinh y - (-e^x\sinh y) = 2e^x \sinh y \ne 0; \tag 5$
we conclude $\vec V(x, y)$ is not a gradient and hence no function $v(x, y)$ such that $\vec V(x, y) = \nabla v(x, y)$ exists. Thus $u(x, y)$ has no harmonic conjugate, and thus cannot be the real part of an analyic function.
If one wants to follow through with the approach suggested by our OP Mophotla, note that
$e^x\cosh(y)+C_1(x)=-e^x\cosh(y)+C_2(y) \tag 6$
yields
$C_2(y) - C_1(x) = 2e^x \cosh y; \tag 7$
taking $x = 0$ in (7),
$C_2(y) = C_1(0) + 2\cosh y, \tag 8$
whereas with $y= 0$ we find
$C_1(x) = -2e^x + C_2(0); \tag 9$
thus,
$2\cosh y + 2e^x + (C_1(0) - C_2(0)) = 2e^x \cosh y; \tag{10}$
since $e^x\cosh y \ne 0$ for any $x, y$, we may write
$2e^{-x} + \dfrac{2}{\cosh y} + \dfrac{C_1(0) - C_2(0)}{e^x \cosh y} = 2; \tag{11}$
now letting $x, y \to \infty$ together, we see that $e^x, \cosh y, e^x \cosh y \to \infty$, and (11) yields the contradiction
$0 = 2; \tag{12}$
thus functions such as $C_1(x)$, $C_2(y)$ cannot exist, and this finishes off the proof, according to our OP Mophotla's suggested method, that $u(x, y)$ is not the real part of an analytic function.
Best Answer
If $u$ is the real part of a holomorphic function, $u$ is harmonic (and this follows from Cauchy-Riemann equations).
Thus, a necessary condition for $y$ to be the real part of a holomorphic function is that $$ \Delta u=0\\ 2+2b=0\\ b=-1 $$
Conversely, a harmonic function in all of $\mathbb{R}^2$ is the real part of an holomorphic function (which in our case you can determine explicitly), and we are done