It is a problem in my homework. I have worked out the MLE of $\sigma_1$ and $\sigma_2$. $\sigma_1^{MLE} = \frac{\sum_{i=1}^{n}X_iI_{(Xi>0})}{n}$+$\frac{\sqrt{\sum_{i=1}^{n}X_iI_{(Xi>0)}} }{\sqrt{n}} *\frac{\sqrt{\sum_{i=1}^{n}-X_iI_{(Xi<0)}} }{\sqrt{n}}$
I wonder how to deal with the second term when n goes to infinty.
First of all, you found MLE for $\sigma_1$ right! With $n\to\infty$, the term $$ \sqrt{\frac{\sum_{i=1}^n X_iI_{(X_i>0)}}{n} }\cdot\sqrt{\frac{\sum_{i=1}^n -X_iI_{(X_i<0)}}{n} } $$ goes in probability due to LLN to a constant $$ \sqrt{\mathbb E\bigl[X_1I_{(X_1>0)}\bigr]}\cdot \sqrt{\mathbb E\bigl[X_1I_{(X_1>0)}\bigr]}. $$ Here $$ \mathbb E\bigl[X_1I_{(X_1>0)}\bigr]=\int_0^\infty x\cdot c\exp(-x/\sigma_1)\,dx = c\sigma_1^2 = \frac{\sigma_1^2}{\sigma_1+\sigma_2} $$ and $$ \mathbb E\bigl[-X_1I_{(X_1<0)}\bigr]=\int_{-\infty}^0 -x\cdot c\exp(x/\sigma_2)\,dx = c\sigma_2^2 = \frac{\sigma_2^2}{\sigma_1+\sigma_2} $$ So $$ \sqrt{\frac{\sum_{i=1}^n X_iI_{(X_i>0)}}{n} }\cdot\sqrt{\frac{\sum_{i=1}^n -X_iI_{(X_i<0)}}{n} } \xrightarrow{p} \frac{\sigma_1\sigma_2}{\sigma_1+\sigma_2}. $$ And MLE for $\sigma_1$ is consistent: $$ \sigma_1^{MLE}\xrightarrow{p}\frac{\sigma_1^2}{\sigma_1+\sigma_2} +\frac{\sigma_1\sigma_2}{\sigma_1+\sigma_2} = \sigma_1 $$
For asymptotic distribution of pair of estimates $(\sigma_1^{MLE},\sigma_2^{MLE})$ you probably need to use multivariate delta-method.