Since $\mathbb{P}(X+Y>0 \vert Y<0) = \frac{\mathbb{P}(X+Y>0 \land Y<0)}{\mathbb{P}(Y<0)}$, and the denominator being a normal CDF, the real problem is to compute $\mathbb{P}(X+Y>0 \land Y<0)$.
Note, that the pair $(U,V) = (X+Y,Y)$ has bivariate normal distribution with means $(2 \mu, \mu)$, variances $(2 \sigma^2, \sigma^2)$ and correlation $\rho = \frac{1}{\sqrt{2} \sigma^2} \mathbb{E}((X+Y-2\mu)(Y-\mu)) = \frac{1}{\sqrt{2} \sigma^2} \mathbb{E}((Y-\mu)^2) = \frac{1}{\sqrt{2}}$.
Thus we seek $\mathbb{P}(U > 0 \land V < 0) = \mathbb{P}(V < 0) - \mathbb{P}(U < 0 \land V < 0)$. The last part is the CDF of the bivariate normal, which has no closed form for generic $\mu$ and $\sigma$. So
$ \mathbb{P}(X+Y>0 \vert Y<0) = 1 - \frac{\rm CDF_{U,V}(0,0)}{ \rm CDF_{V}(0)} $ Now using $\rm CDF$ of standard normal and standard bi-normal with correlation $\rho=\frac{1}{\sqrt{2}}$: $ \mathbb{P}(X+Y>0 \vert Y<0) = 1 - \frac{\rm CDF_{\mathcal{BN}(\rho)}(-\sqrt{2}\frac{\mu}{\sigma},-\frac{\mu}{\sigma})}{ \Phi(-\frac{\mu}{\sigma}) } $
The following plot shows $\mathbb{P}(X+Y>0 \vert Y<0)$ as a function of $\frac{\mu}{\sigma}$:
