One can use Mellin transforms to tackle this integral; in this case one obtains series involving hypergeometric functions. Here is a very brief summary of the process. I plan to complete the answer with more details, and with an asymptotic series for large $a$ soon.
Setting $$f(x) = \frac{\sin(ax)}{x}$$ and $$g(x) = I_1(bx) K_1(cx),$$ one has the Mellin transforms
$$ F(s) = a^{1-s} \cos \left( \frac{\pi s}{2}\right) \Gamma(s-1) $$ and
$$ G(s) = 2^{s-2} \frac{b}{c^{s+1}}\ \Gamma(s/2)\ \Gamma(1+s/2)\ {}_2 F_1\! \left[ \frac{s}{2}, \frac{s+2}{2}; 2 ; \frac{b^2}{c^2} \right].$$
The Parseval theorem for Mellin transforms gives
$$ I = \frac{1}{2 \pi i} \int_{k-i\infty}^{k+i\infty} ds\ F(1-s)\ G(s), $$ where $0<k<1.$ The contour can be moved along the real axis to plus or minus infinity, picking up residues from the poles as you go.
Simple poles are found at $s=2k+1, k=0,1,2,\cdots,$ and displacing the contour over these gives a series useful for $a<1$, given by
$$ I(a,b,c) = \frac{b}{2} \sum_{k=0}^\infty \frac{a^{2k+1}}{(2k+1)!} \frac{(-1)^k}{c^{2k+2}} 4^k \left(k+\frac{1}{2}\right)\Gamma^2\left( k+\frac{1}{2}\right) {}_2 F_1\! \left[ k+\frac{1}{2}, k+\frac{3}{2} ; 2 ; \frac{b^2}{c^2} \right] .$$
By computing the hypergeometric function for several values of $k$, I find that it can be expressed in each case as a sum of elliptic integrals of the first and second kind, times polynomials, reminiscent of your answer above. (This was done with Mathematica.)
EDIT:
Alternatively, one can displace the contour over the simple poles located at $s = -2 k, k=0,1,2,\cdots,\infty$ to obtain an asymptotic series useful for large $a$, which after some simplification becomes
$$ I = \frac{\pi b}{4 c} \left( 1 + \sum_{k=1} (-1)^k \left(\frac{c}{2 a} \right)^{2 k} \frac{ (2 k)!}{k!^2}\ {}_2F_1\!\left[ -k, -k+1;2;\frac{b^2}{c^2} \right] \right). $$ Note that this series, being asymptotic, must be used with care; i. e. if too many terms are used it will diverge.
In this case the hypergeometric series terminate, and can be evaluated as polynomials in $b^2/c^2$. This can lead to relatively simple expressions useful for large $a$.
