$u$ need not be compactly supported.

Take any $\varphi\in C_0^{\infty}$, $\varphi\not\equiv 0$, and let $\widehat{u}=\widehat{\varphi}^{1/2}$ be any non-holomorphic (but measurable) square root. (Typically, $\widehat{u}$ will fail to be holomorphic automatically, when $\widehat{\varphi}$ has zeros.) Clearly $\widehat{u}\in L^2$, since $\widehat{\varphi}$ is a Schwartz function.

Then $(u*u)\widehat{\:\:\:}=\widehat{u}\,^2=\widehat{\varphi}$, so $u*u=\varphi$ is compactly supported, but $u$ isn't because its Fourier transform is not entire.

$u$ need not be compactly supported.

Take any $\varphi\in C_0^{\infty}$, $\varphi\not\equiv 0$, and let $\widehat{u}=\widehat{\varphi}^{1/2}$ be any non-holomorphic (but measurable) square root. (Typically, $\widehat{u}$ will fail to be holomorphic automatically, when $\widehat{\varphi}$ has zeros.) Clearly $\widehat{u}\in L^2$, since $\widehat{\varphi}$ is a Schwartz function.

Then $(u*u)\widehat{\:\:\:}=\widehat{u}\,^2=\widehat{\varphi}$, so $u*u=\varphi$ is compactly supported, but $u$ isn't because its Fourier transform is not entire.

We can also produce explicit examples in this fashion, if we start out with a (non-smooth) $\varphi$ whose Fourier transform we can compute. For example, $u(x)=\log |(x-1)/(x+1)|$ works; this comes from a triangular $\varphi$.