Rigorous formulation
When mathematical physicists work with point interactions, the “delta” in the potential is only a formal shorthand. What they actually mean are the corresponding boundary conditions $$\psi'(0+) - \psi'(0-) = -\lambda \, \psi(0) \: .$$
The Hamiltonian in your question is then rigorously defined as:
\begin{gather*}
\mathscr D_H = \big\{ \;
\psi \in W^{1,2}(\mathbb R) \cap W^{2,2}(\mathbb R \setminus \{0\})
\;\; \big| \;\;
\psi'(0+) - \psi'(0-) = -\lambda \, \psi(0)
\; \big\} \: ,
\\[10pt]
\big( H \psi \big)(x) = -\psi''(x)
\quad \text{ for a.e. } \quad x \in \mathbb R \setminus \{0\} \: ,
\end{gather*}
where $W^{k,p}$ are the Sobolev spaces of $k$-times weakly differentiable $L^p$ functions.
Self-adjointness
You can check that this Hamiltonian is indeed self-adjoint:
\begin{align*}
( H\psi, \varphi)
&= \int_{-\infty}^{+\infty} \!\!\! -\overline{\psi''} \, \varphi
\;=\; \int_{-\infty}^{0} \!\!\! -\overline{\psi''} \, \varphi
\;+\; \int_{0}^{+\infty} \!\!\! -\overline{\psi''} \, \varphi
\\[5pt]
&= \big[-\overline{\psi'}\varphi \big]_{-\infty}^{0}
+ \big[ -\overline{\psi'}\varphi\big]_{0}^{+\infty}
\;+\; \int_{-\infty}^{0} \!\!\! \overline{\psi'} \, \varphi'
\;+\; \int_{0}^{+\infty} \!\!\! \overline{\psi'} \, \varphi'
\\[5pt]
&= \big[-\overline{\psi'}\varphi \big]_{-\infty}^{0}
+ \big[ -\overline{\psi'}\varphi\big]_{0}^{+\infty}
+ \big[-\overline{\psi}\varphi' \big]_{-\infty}^{0}
+ \big[ -\overline{\psi}\varphi'\big]_{0}^{+\infty}
+ \int_{-\infty}^{+\infty} \!\!\! -\overline\psi \, \varphi''
\\[5pt]
&= \big[-\overline{\psi'}\!\varphi -\overline{\psi}\varphi' \big]_{-\infty}^{+\infty} - \lambda \, \overline\psi(0) \, \varphi(0) + \overline\psi(0) \, \big( \varphi'(0+) - \varphi'(0-) \big)
+ \!\int_{-\infty}^{+\infty} \!\!\!\!\! -\overline\psi \, \varphi''
\end{align*}
which is equal to $(\psi, H\varphi)$ exactly iff $\varphi \in \mathscr D_H$. Really, the terms at $\pm\infty$ are zero, because functions from $W^{1,2}$ vanish at infinity (proof) and the terms at $0\pm$ exactly cancel out if $\varphi$ obeys the boundary condition.
Motivation
This might seem unsatisfactory, because the rigorous formulation sidesteps the “multiplication by delta” operator. Therefore I will hint at a construction that follows the formal notation more closely. Choose your favourite space of test functions $\mathcal D$ (eg. smooth functions with compact support) and the space of continuous linear functionals on it $\mathcal D'$. Since $\mathcal D$ is naturally embedded in $L^2(\mathbb{R})$ and $L^2(\mathbb{R})$ is naturally embedded in $\mathcal D'$, we have the famous Gelfand sandwich $\mathcal D \subset \mathcal H \subset \mathcal D'$. Now we can define two operators. The first one is the distributional second derivative:
$$
\hat \Delta: \mathcal H \to \mathcal D' \: ,
\qquad
\langle \hat \Delta \psi, \varphi \rangle = \int \psi \, \varphi''
\quad \forall \psi \in \mathcal H
\quad \forall \varphi \in \mathcal D \: .
$$
This is exactly how derivatives work on distributions (source). The second operator we define is the operator of multiplication by delta:
$$
\hat \delta: \big\{ \; \psi \in \mathcal H \; \big| \; \psi \text{ is continuous at } 0 \; \big\} \to \mathcal D' \: ,
\;\;
\langle \hat \delta \psi, \varphi \rangle = \psi(0) \, \varphi(0)
\;\, \forall \psi \in \mathcal H
\;\, \forall \varphi \in \mathcal D \: .
$$
Now we can rigorously define the Hamiltonian as $\hat H = -\hat \Delta - \lambda \hat \delta$ which is an operator from a subset of $L^2(\mathbb{R})$ to the distributions $\mathcal D'$. We could ask if there is a restriction of $\hat H$ to a self-adjoint operator on $L^2(\mathbb{R})$. Yes, there is – it is $H = \hat H|_{\mathscr D_H}$, as defined above!
