Yes. Let 
$$ F(g) = \sum_{h \in G} \overline{\chi(h)}\chi(gh).$$
I claim that $F$ is a class function. Indeed,
$$F(s^{-1} g s) = \sum_{h \in G} \overline{\chi(h)}\chi(s^{-1}gsh)=\sum_{h \in G} \overline{\chi(h)}\chi(g(shs^{-1})) = \sum_{h' \in G} \overline{\chi(s^{-1}h's )}\chi(gh') = \sum_{h' \in G} \overline{\chi(h' )}\chi(gh') = F(g).$$

Thus, we can write $F$ as a sum of irreducible characters:
$$F(g) = \sum_{i} a_{\chi_i} \chi_i(g).$$
Let $\rho_{\chi_i}$ be the representation with character $\chi_i$. Then
$$a_{\chi_i} = \frac{1}{|G|} \sum_{g \in G} \overline{\chi_i(g)}F(g) = \frac{1}{|G|} \sum_{g,h} \overline{\chi_i(g)} \overline{\chi(h)}\chi(gh)= \frac{1}{|G|} \sum_{h} \overline{\chi(h)} (\mathrm{Tr}( \rho_{\chi}(h) \sum_{g} \overline{\chi_i(g)}\rho_{\chi}(g) ).  $$
Letting
$$A_i = \sum_{g} \overline{\chi_i(g)}\rho_{\chi}(g),$$
we have $$(*)a_{\chi_i}  = \frac{1}{|G|} \sum_{h} \overline{\chi(h)} (\mathrm{Tr}( \rho_{\chi}(h) A_i).$$ 
I claim $A_i$ commutes with $\rho_{\chi}(h)$ for all $h$-s:
$$A_i \rho_{\chi}(h) =\sum_{g} \overline{\chi_i(g)}\rho_{\chi}(gh) =\sum_{g'} \overline{\chi_i(g'h^{-1})}\rho_{\chi}(g')=\sum_{g'} \overline{\chi_i(h^{-1} g')}\rho_{\chi}(g')=\sum_{g} \overline{\chi_i(g)}\rho_{\chi}(hg)= \rho_{\chi}(h)A_i.$$
Thus, by Schur's lemma, $A_i$ is a multiple of the identity:
$$A_i = c_i I$$
for some constant $c_i$, which can be extracted by taking traces:
$$c_i \cdot \dim(\rho_i) = \mathrm{Tr}(A_i) = \sum_{g \in G}\overline{\chi_i}(g) \chi(g) = 1_{\chi_i = \chi} |G|.$$ 
Thus, wen $\chi_i \neq \chi$ we have $A_i=0$ and so $a_{\chi_i}=0$. If $\chi_i=\chi$, we have from $(*)$
$$a_{\chi_i}=\frac{1}{\dim \rho_{\chi_i}}\sum_{h} \overline{\chi_i}(h) \chi(h) = |G|/\chi(e),$$
as needed.