To (potentially) cut down on the number of unanswered questions, I'll post GEdgar's answer from the comments as a Community Wiki answer.
Since we're assuming $f$ is holomorphic, then $$f(z)=\sum_{n=0}^\infty a_nz^n$$ holds for all $z\in\Bbb C$, where each $a_n\in\Bbb C$. Since differentiation of a power series is done termwise, then $$f''(z)=\sum_{n=0}^\infty\frac{d^2}{dz^2}[a_nz^n]=\sum_{n=2}n(n-1)a_nz^{n-2}=\sum_{n=0}^\infty(n+2)(n+1)a_{n+2}z^n,$$ so since $f+f''$ is identically zero on $\Bbb C$, then $a_n=-(n+2)(n+1)a_{n+2}$ for all $n\geq 0$.
Let's set $C=a_0$ and $A=a_1$. Since $a_n=-(n+2)(n+1)a_{n+2}$ for all $n\geq 0$, then quick inductive arguments show that for all $k\geq 0$ we have $$a_{2k}=\frac{(-1)^kC}{(2k)!}$$ and $$a_{2k+1}=\frac{(-1)^kA}{(2k+1)!}.$$ Consequently, we have
$\begin{eqnarray*}
f(x) & = & \sum_{n=0}^\infty a_nz^n\\
& = & \sum_{k=0}^\infty a_{2k+1}z^{2k+1}+\sum_{k=0}^\infty a_{2k}z^{2k}\\
& = & A\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}z^{2k+1}+C\sum_{k=0}^\infty\frac{(-1)^k}{(2k)!}z^{2k}\\
& = & A\sin z+C\cos z.
\end{eqnarray*}$
As an addendum, your ansatz $A\sin(bz)+C\cos(bz)$ is not a solution to the differential equation, in general. We may as well assume that $b$ isn't a negative number. Otherwise, we can use odd/even properties of sine and cosine to rewrite it as $A_0\sin(b_0z)+C_0\cos(b_0z)$, where $A_0=-A,C_0=C,b_0=-b$. Note that if $g(z)=A\sin(bz)+C\cos(bz)$ for some $A,C,b\in\Bbb C$, then $$g(z)+g''(z)=(1-b^2)g(z).$$ Consequently, if $g$ isn't identically zero (that is, if $A,C$ not both zero), then $g+g''=0$ if and only if $b=1$ (since $b=1$ is the only nonnegative solution to $1-b^2=0$).
