Let $g:(-\infty,0] \to [0,\infty)$ be a $C^2$ strictly decreasing function, $g(0)=0$.
Suppose that there exist a sequence $\lambda_n \to -\infty$ such that $$ g(\frac{x + y}{2}) \le \frac{g(x) + g(y)}{2} \, \,\, \, \text{ whenever } \, \, x,y \le 0 \, \, \, \text{ and } x+y=\lambda_n. $$
Does $\lim_{x \to -\infty}g(x)=+\infty$?
If we were assuming the stronger assumption that $g(\frac{x + y}{2}) \le \frac{g(x) + g(y)}{2}$ whenever $x,y \in (-\infty,0]$, then the answer would positive:
Indeed, full midpoint-convexity implies convexity, and a convex function lies above its tangent. ($g$ have a point with negative derivative, since it is strictly decreasing).
Yes, the conjecture is true. Since $\lambda_n \to -\infty$ we can choose a subsequence $(\lambda_{n_k})$ such that $$ \lambda_{n_k} < 2 \lambda_{n_{k-1}} $$ for all $k$. Choosing $x=0$ and $y= \lambda_{n_k}$ in your condition and the monotonicity of $g$ gives $$ g(\lambda_{n_{k-1}}) \le g\left( \frac 12 \lambda_{n_k}\right) \le \frac 12 g(\lambda_{n_k}) $$ so that $$ g(\lambda_{n_k}) \ge 2 g(\lambda_{n_{k-1}}) \ge 2^2 g(\lambda_{n_{k-2}}) \ge \ldots \ge 2^{k-1} g(\lambda_{n_{1}}) \, . $$