By Andreieff's identity:
$$
{\rm det}(A)=\frac{1}{n!}\int_{(0,\infty)^{n}}
{\rm det}[e^{-\frac{t_k}{2}}t_k^{\lambda_i-\frac{1}{2}}]_{1\le i,k\le n}\ \times\ 
{\rm det}[e^{-\frac{t_k}{2}}t_k^{\mu_j-\frac{1}{2}}]_{1\le k,j\le n}\ \ dt_1\cdots dt_n
$$
$$
=\frac{1}{n!}\int_{(0,\infty)^{n}}
\left(\prod_{l=1}^{n} \frac{e^{-t_l}}{t_l}\right)\times 
{\rm det}[t_k^{\lambda_i}]_{1\le i,k\le n}\ \times\ 
{\rm det}[t_k^{\mu_j}]_{1\le k,j\le n}\ \ dt_1\cdots dt_n\ .
$$
By the identity, e.g., in 
https://mathoverflow.net/questions/277655/reference-for-exponential-vandermonde-determinant-identity/277666#277666
the two determinants in the integrand have the same sign (write $t_l=e^{\alpha_l}$ to see the usual expression of the Harish-Chandra-Itzykson-Zuber integral). It is then easy to conclude that ${\rm det}(A)>0$ and so $A$ is nonsingular.