It's true.
Since $A$ acts irreducibly, so does $\mathfrak{a}$, so the latter is reductive. Write $\mathfrak{a}=\mathfrak{z}\oplus\mathfrak{s}$, with $\mathfrak{z}$ its center and $\mathfrak{s}=[\mathfrak{a},\mathfrak{a}]$ being semisimple.
Assume that $A$ has no element with a nonzero real eigenvalue. This implies that $\mathfrak{s}$ has real rank zero (so $[A,A]$ is compact semisimple).
If both $\mathfrak{s}$ and $\mathfrak{z}$ are nonzero, they contain nonzero elements $s$ and $z$. Necessarily $s$ has a nonzero eigenvalue (in $i\mathbf{R}$), say $i$ up to rescaling, and also $z$ being central and centralizing an irreducible representation, we have a decomposition of $\mathbf{C}^d$ into two eigenspaces $V_1\oplus V_2$, on which the eigenvalues of $z$ are $a+ib$ and $a-ib$. Since $s$ commutes with $z$, the eigenspace of $s$ for the eigenvalue $i$ meets at least one of these two spaces. Therefore either $z+bs$ or $z-bs$ has the real eigenvalue $a$. If $a\neq 0$, we are done.
The argument works unless every element of $\mathfrak{z}$ has only eigenvalues in $i\mathbf{R}$, but in this case, $A$ has a compact closure and this is not compatible with the density assumption.