First of all, I think you should really absorb the stray $u$ on the left-hand side by either $Vu$ or $\lambda u$, so consider
$$
-u''+Vu =\lambda u ,
$$
under the same assumptions. Next, instead of talking about an exponentially decaying solution (whose existence you'd have to prove first), I think it would be much better to take the unique (up to a factor) solution $u(x,\lambda)\in L^2(0,\infty)$, which is guaranteed to exist for $\lambda\notin\mathbb R$ by basic theory; in more technical terms, a bounded $V$ is in the limit point case at infinity. (If there is an exponentially decaying solution, then it's of course the same $u$.)

Then the quotient $m(\lambda)=u'(x,\lambda)/u(x,\lambda)$ defines the Titchmarsh-Weyl $m$ function of the problem on $[x,\infty)$, with Dirichlet boundary conditions $v(x)=0$ at $x$. This has a holomorphic extension to the complement of the spectrum, so in your situation is defined for all small enough real $\lambda$. Its asymptotics have been studied extensively, and the answer to your question is that
$$
m(\lambda) = -\sqrt{-\lambda} + \textrm{error term}, \quad \lambda\to -\infty ,
$$
with rather precise information on the error. In particular, indeed $m\to-\infty$, as you suspected.

See reference 271 here for much additional information (the asymptotic formula, as $\lambda\to -\infty$ is right at the beginning, in the abstract).