First of all, the functions of operators are always to be understood as a Taylor expansion. In your case:
\begin{equation}
e^{\lambda a_+} = \sum_{n=0}^{+\infty} \frac{(\lambda a_+)^n}{n!}.
\end{equation}
Let is now consider operators $A$ and $B$ whose commutator is a complex number (i.e. not an operator): $[A, B] = c$. And let us take an infinitely differentiable function $f(x)$. Then we can study the following commutator:
\begin{equation}
[A, f(B)] = \sum_{n=0}^{+\infty}\frac{f^{(n)}}{n!}[A, B^n]
\end{equation}
Using the commutator of $A$ and $B$ we can derive:
\begin{equation}
AB^n = (AB - BA + BA)B^{n-1} = cB^{n-1} + BAB^{n-1} = 2cB^{n-1} + B^2AB^{n-2} = ... = ncB^{n-1} + B^nA,
\end{equation}
that is
\begin{equation}
[A, B^n] = ncB^{n-1}.
\end{equation}
Substituting this into the expansion above we have
\begin{equation}
[A, f(B)] = \sum_{n=1}^{+\infty}\frac{f^{(n)}}{n!}ncB^{n-1} =
c\sum_{n=1}^{+\infty}\frac{f^{(n)}}{(n-1)!}B^{n-1} = c\frac{\partial}{\partial B}f(B).
\end{equation}
(I took into account that $[A,1]=0$.)
This is a very useful formula, which btw is a particular case of more general Baker-Campbell-Hausdorff formula.
Going back to your problem, since $[a_-, a_+] = 1$, we have
\begin{equation}
[a_-, e^{\lambda a_+}] = \lambda e^{\lambda a_+} \Longrightarrow a_-e^{\lambda a_+} = e^{\lambda a_+}(\lambda + a_-).
\end{equation}
We now can prove that $\psi_\lambda$ is an eigenfunction of $a_-$ in a straightforward way:
\begin{equation}
a_-\psi_\lambda = a_-e^{\lambda a_+}\psi_0 = e^{\lambda a_+}(\lambda + a_-)\psi_0 = \lambda e^{\lambda a_+}\psi_0 = \lambda \psi_\lambda,
\end{equation}
since $a_-\psi_0 = 0$.
