The other answer notwithstanding, this is not "just a formal solution", it is the actual solution to the Dirichlet problem in the unit disk.
Let's say $f\in C(\Bbb T)$ just to keep things simple (that is, $f$ is continuous and has period $2\pi$). We want a function $u(r,t)$ which is continuous in the closed unit disk $\overline D$, is harmonic in $D$, and which equals $f$ on the boundary.
The notation is simpler if we use the "complex" form of Fourier series. Let $$c_n=\hat f(n)\quad(n\in\Bbb Z).$$The $c_n$ are certainly bounded, so the series $$u(r,t)=\sum_{n=-\infty}^\infty c_nr^{|n|}e^{int}$$converges uniformly on compact subsets of the unit disk.
Since $z^n$ and $\left(\overline z\right)^n$ are analytic it follows that the partial sums are harmonic. Since a uniform limit of harmonic functions is harmonic this shows that $u$ is harmonic in the (open) disk.
To answer a question you ask in a comment, the Fourier series of $f$ need not converge. But that doesn't matter. Look up "Abel summability" somewhere; it's very well known that the Fourier series of $f$ is uniformmly Abel summable to $f$; this means precisely that $u(r,t)\to f(t)$ uniformly as $r\to 1$.
About convergence to $f$ at the boundary: Say $P_r$ is the Poisson kernel. It's well known and not that hard to prove that the convolution $P_r*f$ tends to $f$ uniformly. It's also well known that $$P_r(t)=\sum_{n=-\infty}^\infty r^{|n|}e^{int}$$(you can calculate that sum explicitly, since it's just the sum of two geometric series), which makes it clear that $$P_r*f(t)=u(r,t).$$
Come to think of it, since the title asks about harmonic functions without mentioning boundary values, it may be appropriate to point out that in fact any function $u(r,t)$ harmonic in the open unit disk is given by a series as above for some choice of coefficients $c_n$.
Outline of proof: If $u$ is a real-valued harmonic function in the open disk we know there exists a real-valued harmonic function $v$ such that $u+iv$ is analytic. This shows that if $u$ is any harmonic function in the disk there exist analytic functions $f$ and $g$ such that $$u=f+\overline g.$$Now $f$ and $g$ are both given by power series...
Let $u$ a (real) harmonic function on the exterior of a disc of radius $k$ centered at the origin and continuous on the boundary circle.
By Green's theorem, one gets that there is a constant $B$ (not depending on $R$) given by $2\pi B=\int_{0}^{2\pi}R\frac{\partial u}{\partial r}(R,\theta)d\theta, R>k$, which immediately implies that the harmonic function $u(r, \theta)-B \log r$ has a harmonic conjugate, so it is the real part of an analytic function on the disc exterior.
This implies that $\int_{0}^{2\pi}u(R,\theta)d\theta=A+2\pi B \log R, R >k$
(if $2u=2\Re f= f+\bar f$ the Laurent series of $f$ gives that $\int_{0}^{2\pi}u(R,\theta)d\theta$ is constant in $R$ since only the free term in the Laurent series has non zero integral on a circle and then apply the above to $u-B \log r$).
All the above is general stuff about harmonic function on an annulus or disc exterior; applied to the OP problem, we get that $A=B=0$ by taking $R \to \infty$ and using that $\lim_{r\to\infty} u(r,\theta) = 0$; but now taking $R \to 1$ we get the neceesary condition that $\int_{0}^{2\pi}f(\theta)d\theta =0$. The condition is also sufficient as noted in the OP.
Best Answer
I don't know what you did exactly in your computation process. At any rate the "eigenvalue" $n=0$ is a double eigenvalue, and besides the constant solution there is an additional solution, namely $u(r,\theta):=\log r$. This should give you enough manoeuvring space to satisfy the boundary conditions with a rotationally symmetric solution.