Sometimes you can

Since there seem to be no hard results along these lines on the literature, I decided to have a look at the two shapes in the question and see if there are some relatively accessible results and happily, as it turns out, there are. In particular:

The isospectral surfaces $D_1$ and $D_2$ in the question are acoustically distinguishable: there exist points $x_1^*\in D_1$ and $x_2^*\in D_2$ such that no point $x_2\in D_2$ has the same timbre $\tau^{(2)}(x_2)$ as $\tau^{(1)}(x_1^*)$ and no point $x_1\in D_1$ has the same timbre $\tau^{(1)}(x_1)$ as $\tau^{(2)}(x_2^*)$.

Drummers hitting those drums can therefore prove conclusively which drum they are using by an appropriate choice of the drumming point.

To explore the eigenfunctions, I used the Helmholtz solver detailed in this Mathematica.SE thread, which when implemented directly returns eigenvalues for the two surfaces within a few parts in $10^{-4}$ of each other. The first few eigenfunctions for the two domains look something like this:

What really matters here is the nodes, shown as thick black lines, though of course the timbres $\tau(x)$ as I defined them contain more information in terms of the relative weights of the eigenfunctions when they're not zero. To begin with, already by the fourth eigenfunction we're able to 'count' the shape of the drum as in Carlo Beenakker's answer (i.e. if you have access to the number of nodal domains for that eigenvalue, then you can immediately tell them apart). The nodes, however, also give quick and clear examples of the $x^*$s we seek.

In particular, consider the nodes of $\varphi_2$ and $\varphi_4$ for the two domains, shown in black and red respectively,

and similarly the nodes of $\varphi_2$ and $\varphi_5$:

Specifically, note that the nodes of $\varphi_2$ and $\varphi_4$ cross for $D_2$ but not for $D_1$, so a drummer hitting the drum in that position will produce a sound with strong components of $\lambda_1, \lambda_3, \lambda_5, \lambda_6, \lambda_7$, and so on, but with no Fourier component along the frequencies $\lambda_2$ and $\lambda_4$; this is completely impossible for a drummer using $D_1$.

Similarly, a drummer using $D_1$ can produce a sound with no $\lambda_2$ or $\lambda_5$ component by hitting the intersection of the corresponding nodes, and such a sound is impossible to produce using $D_2$.

The Mathematica code used to produce this answer is available here.

It seems that the full question, however, is still open: whether you always can, or whether you sometimes can't. That is, it would be interesting to know whether there exist non-isometric surfaces which are not only isospectral, but also acoustically indistinguishable in the sense laid out in the question.

Sometimes you can

Since there seem to be no hard results along these lines on the literature, I decided to have a look at the two shapes in the question and see if there are some relatively accessible results and happily, as it turns out, there are. In particular:

The isospectral surfaces $D_1$ and $D_2$ in the question are acoustically distinguishable: there exist points $x_1^*\in D_1$ and $x_2^*\in D_2$ such that no point $x_2\in D_2$ has the same timbre $\tau^{(2)}(x_2)$ as $\tau^{(1)}(x_1^*)$ and no point $x_1\in D_1$ has the same timbre $\tau^{(1)}(x_1)$ as $\tau^{(2)}(x_2^*)$.

Drummers hitting those drums can therefore prove conclusively which drum they are using by an appropriate choice of the drumming point.

To explore the eigenfunctions, I used the Helmholtz solver detailed in this Mathematica.SE thread, which when implemented directly returns eigenvalues for the two surfaces within a few parts in $10^{-4}$ of each other. The first few eigenfunctions for the two domains look something like this:

What really matters here is the nodes, shown as thick black lines, though of course the timbres $\tau(x)$ as I defined them contain more information in terms of the relative weights of the eigenfunctions when they're not zero. To begin with, already by the fourth eigenfunction we're able to 'count' the shape of the drum as in Carlo Beenakker's answer (i.e. if you have access to the number of nodal domains for that eigenvalue, then you can immediately tell them apart). The nodes, however, also give quick and clear examples of the $x^*$s we seek.

In particular, consider the nodes of $\varphi_2$ and $\varphi_4$ for the two domains, shown in black and red respectively,

and similarly the nodes of $\varphi_2$ and $\varphi_5$:

Specifically, note that the nodes of $\varphi_2$ and $\varphi_4$ cross for $D_2$ but not for $D_1$, so a drummer hitting the drum in that position will produce a sound with strong components of $\lambda_1, \lambda_3, \lambda_5, \lambda_6, \lambda_7$, and so on, but with no Fourier component along the frequencies $\lambda_2$ and $\lambda_4$; this is completely impossible for a drummer using $D_1$.

Similarly, a drummer using $D_1$ can produce a sound with no $\lambda_2$ or $\lambda_5$ component by hitting the intersection of the corresponding nodes, and such a sound is impossible to produce using $D_2$.

The Mathematica code used to produce this answer is available here.

It seems that the full question, however, is still open: whether you always can, or whether you sometimes can't. That is, it would be interesting to know whether there exist non-isometric surfaces which are not only isospectral, but also acoustically indistinguishable in the sense laid out in the question.

#You sometimes can

Since there seem to be no hard results along these lines on the literature, I decided to have a look at the two shapes in the question and see if there are some relatively accessible results; as it turns out, there are. In particular:

The isospectral surfaces $D_1$ and $D_2$ in the question are acoustically distinguishable: there exist points $x_1^*\in D_1$ and $x_2^*\in D_2$ such that no point $x_2\in D_2$ has the same timbre $\tau^{(2)}(x_2)$ as $\tau^{(1)}(x_1^*)$ and no point $x_1\in D_1$ has the same timbre $\tau^{(1)}(x_1)$ as $\tau^{(2)}(x_2^*)$.

Drummers hitting those drums can therefore prove conclusively which drum they are using by an appropriate choice of the drumming point.

To see how, I used the Helmholtz solver detailed in this Mathematica.SE thread, which when implemented directly returns eigenvalues for the two surfaces within a few parts in $10^{-4}$ of each other. The first few eigenfunctions for the two domains look something like this:

What really matters here is the nodes, which give quick and clear examples of the $x^*$s we seek. In particular, consider the nodes of $\varphi_2$ and $\varphi_4$ for the two domains, shown in black and red respectively,

and similarly the nodes of $\varphi_2$ and $\varphi_5$:

In particular, note that the nodes of $\varphi_2$ and $\varphi_4$ cross for $D_2$ but not for $D_1$, so a drummer hitting the drum in that position will produce a sound with strong components of $\lambda_1, \lambda_3, \lambda_5, \lambda_6, \lambda_7$, and so on, but with no Fourier component along the frequencies $\lambda_2$ and $\lambda_4$; this is completely impossible for a drummer using $D_1$.

Similarly, a drummer using $D_1$ can produce a sound with no $\lambda_2$ or $\lambda_5$ component by hitting the intersection of the corresponding nodes, and such a sound is impossible to produce using $D_2$.

The Mathematica code used to produce this answer is available here.

It seems that the full question, however, is still open: whether you always can, or whether you sometimes can't. That is, it would be interesting to know whether there exist non-isometric surfaces which are not only isospectral, but also acoustically indistinguishable in the sense laid out in the question.

Sometimes you can

Since there seem to be no hard results along these lines on the literature, I decided to have a look at the two shapes in the question and see if there are some relatively accessible results and happily, as it turns out, there are. In particular:

The isospectral surfaces $D_1$ and $D_2$ in the question are acoustically distinguishable: there exist points $x_1^*\in D_1$ and $x_2^*\in D_2$ such that no point $x_2\in D_2$ has the same timbre $\tau^{(2)}(x_2)$ as $\tau^{(1)}(x_1^*)$ and no point $x_1\in D_1$ has the same timbre $\tau^{(1)}(x_1)$ as $\tau^{(2)}(x_2^*)$.

Drummers hitting those drums can therefore prove conclusively which drum they are using by an appropriate choice of the drumming point.

To explore the eigenfunctions, I used the Helmholtz solver detailed in this Mathematica.SE thread, which when implemented directly returns eigenvalues for the two surfaces within a few parts in $10^{-4}$ of each other. The first few eigenfunctions for the two domains look something like this:

What really matters here is the nodes, shown as thick black lines, though of course the timbres $\tau(x)$ as I defined them contain more information in terms of the relative weights of the eigenfunctions when they're not zero. To begin with, already by the fourth eigenfunction we're able to 'count' the shape of the drum as in Carlo Beenakker's answer (i.e. if you have access to the number of nodal domains for that eigenvalue, then you can immediately tell them apart). The nodes, however, also give quick and clear examples of the $x^*$s we seek. 

In particular, consider the nodes of $\varphi_2$ and $\varphi_4$ for the two domains, shown in black and red respectively,

and similarly the nodes of $\varphi_2$ and $\varphi_5$:

Specifically, note that the nodes of $\varphi_2$ and $\varphi_4$ cross for $D_2$ but not for $D_1$, so a drummer hitting the drum in that position will produce a sound with strong components of $\lambda_1, \lambda_3, \lambda_5, \lambda_6, \lambda_7$, and so on, but with no Fourier component along the frequencies $\lambda_2$ and $\lambda_4$; this is completely impossible for a drummer using $D_1$.

Similarly, a drummer using $D_1$ can produce a sound with no $\lambda_2$ or $\lambda_5$ component by hitting the intersection of the corresponding nodes, and such a sound is impossible to produce using $D_2$.

The Mathematica code used to produce this answer is available here.

It seems that the full question, however, is still open: whether you always can, or whether you sometimes can't. That is, it would be interesting to know whether there exist non-isometric surfaces which are not only isospectral, but also acoustically indistinguishable in the sense laid out in the question.