I was given that that $f_{X|Y}(x|y=1)=\lambda_1e^{-\lambda_1 x}$ for $x\geq 0$ and $f_{X|Y}(x|y=-1)=\lambda_2e^{\lambda_2 x}$ for $x\leq 0$. $X$ is continuous. $Y$ is discrete and can only take on those two values, $1$ and $-1$, with probability $p$ and $1-p$ respectively. $\lambda_1$ and $\lambda_2$ are positive values. I am trying to get the marginal PDF, $f_x(x)$ for when $x\gt 0$ and for when $x\lt 0 $.
I got $f_x(x)$ = $\sum_{y}P(Y)f_{X|Y}(x|y) = p\lambda_1e^{-\lambda_1 x} +(1-p)\lambda_2e^{\lambda_2 x}$, but after this , I am not sure how to interpret it for $x\gt 0$ and $x\lt 0 $. If I plot $f_{X|Y}(x|1)=\lambda_1e^{-\lambda_1 x}$ for $x\geq 0$, $f_{X|Y}(x|-1)=\lambda_2e^{\lambda_2 x}$ for $x\leq 0$, and $\lambda_1e^{-\lambda_1 x} +\lambda_2e^{\lambda_2 x}$ on the same graph, I obtain the following: plot for a fixed $\lambda$, but even visualizing it doesn't help me understand it any better.
To better keep track of the conditional statements you can rewrite the conditional densities as functions over $\mathbb{R}$ using indicator functions, so $$ f_{X|Y}(x|y=1) = \lambda_1e^{-\lambda_1 x}\mathbb{1}_{x \geq 0}, \qquad f_{X|Y}(x|y=-1)=\lambda_2e^{\lambda_2 x}\mathbb{1}_{x < 0}, $$ we haven't added any new information, but it might help to keep track of everything. So now when you carry out your marginalisation you have $$ \begin{align} f_X(x) &= \sum_{y \in \{-1, 1\}} f_Y(y)f_{X|Y}(x|y) \\ &= p\lambda_1e^{-\lambda_1 x}\mathbb{1}_{x \geq 0} + (1-p)\lambda_2 e^{\lambda_2 x}\mathbb{1}_{x < 0}, \end{align} $$ again nothing new, but hopefully now it should be just a little clearer how to interpret it, so you could also rewrite the density as $$ f_X(x) = \begin{cases} p\lambda_1 e^{-\lambda_1 x} & x \geq 0 \\ (1-p)\lambda_2 e^{\lambda_2 x} & \mbox{otherwise}. \end{cases} $$
So on the sets $\{ x < 0 \}$ and $\{ x > 0 \}$ you are going to have a graph that looks like the density function of an exponential random variable weighted by the probabilities $p$ or $(1-p)$ and you will have a discontinuity at $0$.
A good thing to do next for your own understanding is to probably rewrite the density function in terms of the absolute value of $x$, $| x |$, and see if you can realise the Laplace distribution as a special case of this setup.