This is not a complete answer, but it's a bit long to fit in a comment, so I'm posting it here in case it's useful to someone.

Let's denote:

$$
I(p, q, \lambda; n)
:=
\int_0^{2 \pi} e^{p \cos (\lambda \tau) + q \cos ((1 - \lambda) \tau)}
\cos (n \tau) \frac{\mathrm d \tau}{2 \pi}
$$

There's a symmetry in this integral of the form:

$$
I(p, q, \lambda; n)
=
I(q, p, 1 - \lambda; n)
$$

Also note that the integrand is even, i.e. $f(-\tau) = f(\tau)$.

Now, as mentioned in the OP, we have:

$$
I(p, 0, 1; n)
=
I(0, p, 0; n)
=
I_n(p)
$$

Actually, a slight generalization for $q \neq 0$ gives us:

$$
I(p, q, 1; n)
=
I(q, p, 0; n)
=
e^q I_n (p)
$$

Due to the symmetry property above, taking $\lambda = 1/2$ we actually obtain a rather simple result:

\begin{align*}
I(p, q, \frac{1}{2}; n)
=&
I(q, p, \frac{1}{2}; n)
=
\frac{1}{2 \pi}
\int_0^{2\pi}
\mathrm d\tau\,
e^{(p + q)\cos(\tau / 2)}
\cos (n\, \tau)\\
=&
\frac{1}{\pi}
\int_0^{\pi}
\mathrm dy\,
e^{(p + q)\cos y}
\cos (2\, n\, y)
=
I_{2n}(p + q)
\end{align*}

A special case (probably of limited interest though) is $q = 0$, and $\lambda = 1 / m$, where $m \in \mathbb N$.
Of course, if $m = 1$ or $m = 2$, we obtain the previous results, namely $I_n(p)$ and $I_{2n}(p)$, so we need to take $m \geq 3$ to obtain anything new.
The integral can then be converted into the following form:

$$
I(p, 0, 1/m; n)
=
\frac{m}{2 \pi}
\int_0^{2 \pi / m}
\mathrm dx\,
e^{p \cos x}
\cos (m\, n\, x)
$$

Unfortunately, a general analytic solution doesn't seem likely.
One value which seems to be solvable though is $m = 4$ (originally found by experimenting in Mathematica); in this case, we write the integral as:

$$
I(p, 0, 1/4; n)
=
\frac{2}{\pi}
\int_0^{\pi/2}
\mathrm dx\,
e^{p \cos x}
\cos (4\, n\, x)
$$

We can then use the $n$ angle expansion, so our integral turns into:

$$
I(p, 0, 1/4; n)
=
\frac{2}{\pi}
\sum\limits_{k\;\mathrm{even}}
(-1)^\frac{k}{2}
\begin{pmatrix}4 n\\ k\end{pmatrix}
\int_0^{\pi/2}
\mathrm dx\,
e^{p \cos x}\,
\cos^{4n - k} x\,
\sin^k x
$$

Let's focus on the integral in the above; using the fact that $4 n - k := 2 \ell$ is even, we can rewrite $\cos^{2\ell} x = (1 - \sin^2 x)^\ell$, expand this using the binomial theorem, and additionally expand $e^u = \cosh u + \sinh u$, so that our final result is a linear combination of integrals of the following form:

$$
\mathcal{I}(\alpha, p)
=
\int_0^{\pi/2}
\mathrm dx\,
\sinh(p \cos x)\,
\sin^\alpha x\\
\mathcal{J}(\alpha, p)
=
\int_0^\pi
\mathrm dx\,
\cosh(p \cos x)\,
\sin^\alpha x
$$

for some integer values of $\alpha$.
Note that the second integral has shifted limits, $[0,\pi]$, instead of $[0,\pi/2]$; this is because the integrand is actually symmetric around $\pi/2$, so we can write $\int_0^{\pi/2}(\cdots) = \frac{1}{2} \int_0^\pi (\cdots)$ (the easiest way to see this is to shift the origin to $\pi/2$, i.e. use the substitution $x \rightarrow x - \pi / 2$; then it's simple to demonstrate that $f(-x) = f(x)$, where $f$ denotes the integrand).

These kinds of integrals are known in the literature, see for instance Gradshteyn and Ryzhik, *7th ed.*, formulas 3.997.1 and 3.997.2:

$$
\mathcal{I}(\alpha, p)
=
\frac{\sqrt \pi}{2}
\left(\frac{2}{p}\right)^\alpha
\Gamma\left(\frac{\alpha + 1}{2}\right)
\mathbf{L}_\frac{\alpha}{2}(p)\\
\mathcal{J}(\alpha, p)
=
\sqrt \pi
\left(\frac{2}{p}\right)^\alpha
\Gamma\left(\frac{\alpha + 1}{2}\right)
I_\frac{\alpha}{2}(p)
$$

where $I_\frac{\alpha}{2}$ is the modified Bessel function of the first kind, and $\mathbf{L}_\frac{\alpha}{2}$ is the modified Struve function, see for instance DLMF, chapter 11.

The final result can be shown to be:

\begin{align*}
I(p, 0, 1/4; n)
=&
\frac{2}{\pi}
\sum\limits_{k\;\mathrm{even}}
\sum\limits_{\ell=0}^{2n - k/2}
\begin{pmatrix} 4n \\ k \end{pmatrix}
\begin{pmatrix} 2n - k / 2 \\ \ell \end{pmatrix}
(-1)^{2 n - \ell}
\left[
\mathcal{I}(4n - 2\ell, p)
+
\frac{1}{2}
\mathcal{J}(4n - 2\ell, p)
\right]
\end{align*}

Evidently, using the symmetry property, this automatically gives us $I(0,p,3/4;n)$ as well.

However, I haven't found anything which works for general $m$; while the procedure itself can more or less be repeated for any even $m$, the integrals $\mathcal{I}$ and $\mathcal{J}$ have different integration limits, which do not have any obvious representations in terms of special functions like the above.